Notes on Differential Equations and Hypergeometric Functions

Notes on diﬀerential equations and hypergeometric functions (NOT for publication)

F.Beukers

21st August 2009 Contents

1 Ordinary linear diﬀerential equations 3 1.1 Diﬀerential equations and systems of equations ...... 3 1.2 Local theory ...... 7 1.3 Fuchsian equations ...... 12 1.4 Riemann-Hilbert correspondence ...... 14 1.5 Fuchsian equations of order two ...... 17

2 Gauss hypergeometric functions 21 2.1 Deﬁnition, ﬁrst properties ...... 21 2.2 Monodromy of the hypergeometric function ...... 24 2.3 Triangle groups ...... 31 2.4 Some loose ends ...... 34

3 Hypergeometric functions nFn−1 36 3.1 Deﬁnition, ﬁrst properties ...... 36 3.2 Monodromy ...... 37 3.3 Hypergeometric groups ...... 39 3.4 Rigidity ...... 41

4 Explicit monodromy 45 4.1 Introduction ...... 45 4.2 Mellin-Barnes integrals ...... 46

2 Chapter 1

Ordinary linear diﬀerential equations

1.1 Diﬀerential equations and systems of equations

A differential field K is a field equipped with a derivation, that is, a map ∂ : K → K which has the following properties,

For all a, b ∈ K we have ∂(a + b) = ∂a + ∂b.

For all a, b ∈ K we have ∂(ab) = a∂b + b∂a.

The subset C := {a ∈ K|∂a = 0} is a subfield of K and is called the field of constants. We shall assume that C is algebraically closed and has characteristic zero. We shall also assume that ∂ is non-trivial that is, there exist a ∈ K such that ∂a 6= 0. Standard examples which will be used in later chapters are C(z), C((z)), C((z))an. They are the field of rational functions, formal Laurent series at z = 0 and Lau- rent series which converge in a punctured disk 0 < |z| < ρ for some ρ > 0. As derivation in these examples we have differentiation with respect to z and the field of constants is C. An ordinary differential equation over K is an equation of the form

n n−1 ∂ y + p1∂ y + ··· + pn−1∂y + pny = 0, p1, . . . , pn ∈ K. A system of n ﬁrst order equations over K has the form

∂y = Ay

t in the unknown column vector y = (y1, . . . , yn) and where A is an n × n-matrix with entries in K.

3 4 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

Note that if we replace y by Sy in the system, where S ∈ GL(n, K), we obtain a new system for the new y,

∂y = (S−1AS + S−1∂S)y. Two n × n-systems with coefficient matrices A, B are called equivalent over K if there exists S ∈ GL(n, K) such that B = S−1AS + S−1∂S. It is well known that a differential system can be rewritten as a system by n−1 putting y1 = y, y2 = ∂y, . . . , yn = ∂ y. We then note that ∂y1 = y2, ∂y2 = y3, . . . , ∂yn−1 = yn and finally, ∂yn = −p1yn − p2yn−1 − ..., −pny1. This can be rewritten as       y1 0 1 0 ··· 0 y1  y2   0 0 1 ··· 0   y2  ∂  .  =  . . . .   .   .   . . . .   .  yn −pn −pn−1 −pn−2 · · · −p1 yn There is also a converse statement.

Theorem 1.1.1 (Cyclic vector Lemma) Any system of linear ﬁrst order differential equations over K is equivalent over K to a system which comes from a diﬀerential equation.

Proof. Let ∂y = Ay be our n × n system. Consider the linear form y = r1y1 + ··· + rnyn with r1, . . . , rn ∈ K. Using the diﬀerential system for the yi we see that ∂y = s1y1 + ··· + snyn, where the si are obtained via

(s1, . . . , sn) = ∂(r1, . . . , rn) + (r1, . . . , rn)A.

By repeated application of ∂ we find for each i elements ri1, . . . , rin ∈ K such i that ∂ y = ri1y1 + ··· + rinyn. Denote the matrix (rij)i=0,...,n−1;j=1,...,n by R. If R is invertible, then (rn1, . . . , rnn) is a K-linear combination of the (ri1, . . . , rin) for i = 0, 1, . . . , n − 1. Hence ∂ny is a K-linear combination of the ∂iy (i = 0, . . . , n − 1). Moreover, when R is invertible, our system is equivalent, via R, to a system coming from a differential equation. So it suffices to show that there exist r1, . . . , rn such that the corresponding matrix R is invertible. Since ∂ is non-trivial we can find x ∈ K such that ∂x 6= 0. Note that the new derivation ∂ := (x/∂x)∂ has the property that ∂x = x, which we may now assume without loss of generality. Let µ be the smallest index such that the matrix (rij)i=0,...,µ;j=1,...,n has K-linear dependent rows for every choice of r1, . . . , rn. We must show that µ = n. n Suppose that µ < n. Denote ri = (ri1, . . . , rin). Choose s0 ∈ K such that n s0,..., sµ−1 are independent. Let t ∈ K be arbitrary. By r0 ∧ · · · ∧ rµ we denote the vector consisting of the determinants of all (µ + 1) × (µ + 1) submatrices of the matrix with rows r0,..., rµ. For any λ ∈ C we have now

F.Beukers: Hypergeometric Functions, preliminary notes 1.1. DIFFERENTIAL EQUATIONS AND SYSTEMS OF EQUATIONS 5

(s0 + λt0) ∧ · · · ∧ (sµ + λtµ) = 0. Expand this with respect to powers of λ. Since we have infinitely many choices for λ the coefficient of every power of λ must be zero. In particular the coefficient of λ. Hence

Xµ s0 ∧ · · · ∧ ti ∧ · · · ∧ sµ = 0. (1.1) i=0 ¡ ¢ m n m Pi m i−j Now put t = x u with m ∈ Z and u ∈ K . Notice that ti = x j=0 j m uj. Substitute this in (1.1), divide by xm, and collect equal powers of m. Since m can be chosen in infinitely many ways the coefficient of each power of m must be zero. In particular the coefficient of mµ is zero. Hence

s0 ∧ · · · ∧ sµ−1 ∧ u0 = 0.

Since u0 can be chosen arbitrarily this implies s0 ∧· · ·∧sµ−1 = 0 which contradicts the minimality of µ. 2 We must also say a few words about the solutions of differential equations. It must be pointed out that in general the solutions lie in a bigger field than K. To this end we shall consider differential field extensions L of K with the property that the field of constants is the same as that of K. A fundamental lemma is the following one.

Lemma 1.1.2 (Wronski) Let f1, . . . , fm ∈ K. There exists a C-linear relation between these function if and only if W (f1, . . . , fm) = 0, where ¯ ¯ ¯ f . . . f ¯ ¯ 1 m ¯ ¯ ∂f . . . ∂f ¯ ¯ 1 m ¯ W (f1, . . . , fm) = ¯ . . ¯ ¯ . . ¯ ¯ m−1 m−1 ¯ ∂ f1 . . . ∂ fm is the Wronskian determinant of f1, . . . , fm.

Proof. If the fi are C-linear dependent, then the same holds for the columns of W (f1, . . . , fm). Hence this determinant vanishes. Before we prove the converse statement we need some observations. First notice that m W (vu1, . . . , vum) = v W (u1, . . . , um) for any v, ui ∈ K. In particular, if we take v = 1/um (assuming um 6= 0) we ﬁnd

m m−1 W (u1, . . . , ur)/um = W (u1/um, . . . , um−1/um, 1) = (−1) W (∂(u1/um), . . . , ∂(um−1/um)).

F.Beukers: Hypergeometric Functions, preliminary notes 6 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

Now suppose that W (f1, . . . , fm) vanishes. By induction on m we show that f1, . . . , fm are C-linear dependent. For m = 1 the statement is obvious. So assume m > 1. If fm = 0 we are done, so we can now assume that fm is not zero. By the remarks made above, the vanishing of W (f1, . . . , fm) implies the vanishing of W (∂(f1/fm), . . . , ∂(fm−1/fm)). Hence, by the induction hypothesis, there exist a1, . . . , am−1 such that a1∂(f1/fm) + ··· + am−1∂(fm−1/fm) = 0. After taking primitives and multiplication by fm on both sides we obtain a linear dependence relation between f1, . . . , fm. 2 Lemma 1.1.3 Let L be a differential extension of K with C as field of constants. Then the solution space in L of a linear equation of order n is a C-vector space of dimension at most n. Proof. It is clear that the solutions form a C-vector space. Consider n + 1 solutions y1, . . . , yn+1 and let W be their Wronskian determinant. Note that the columns of this determinant all satisfy the same K-linear relation given by the differential equation. Hence W ≡ 0. According to Wronski’s lemma this implies that y1, . . . , yn+1 are C-linear dependent. 2 Combination of this Lemma with the Cyclic vector Lemma yields Lemma 1.1.4 Let L be a differential extension of K with C as field of constants. Then the solution space in Ln of an n × n-system of equation ∂y = Ay is a C- vector space of dimension at most n. It is allways possible to find differential extensions which have a maximal set of solutions. Without proof we quote the following theorem. Theorem 1.1.5 (Picard-Vessiot) To any n × n-system of linear differential equations over K there exists a differential extension L of K with the following properties 1. The field of constants of L is C. 2. There is an n-dimensional C-vector space of solutions to the system in Ln. Moreover, if L is minimal with respect to these properties then it is uniquely determined up to differential isomorphism.

Let y1,..., yn be an independent set of solutions to an n × n-system ∂y = Ay. The matrix Y obtained by concatenation of all columns yi is called a fundamental solution matrix. The Wronskian lemma together with the Cyclic vector Lemma imply that det(Y ) 6= 0. Exercise 1.1.6 Let Y be the fundamental solution matrix of an n × n-system ∂y = Ay. Prove that det(Y ) satisfies the first order differential equation ∂y = trace(A)y. Show also that the columns of Y −1 satisfy the system of equations ∂y = −Aty, where At denotes the transpose of A.

F.Beukers: Hypergeometric Functions, preliminary notes 1.2. LOCAL THEORY 7

1.2 Local theory

In this section our diﬀerential ﬁeld will be C((z)). We shall denote the derivation d d dz by D and the derivation z dz by θ.

Exercise 1.2.1 Prove by induction on r the following operator identity. For any r ∈ N zrDr = θ(θ − 1) ··· (θ − r + 1). Prove for any m, θ(zmf(z)) = zm(θ + m)f(z).

Consider the linear diﬀerential equation of order n,

n (n−1) D y + p1(z)D y + ··· + pn−1(z)Dy + pn(z)y = 0, (1.2) with pi ∈ C((z)). If z = 0 is not a pole of any pi it is called a regular point of (1.2), otherwise it is called a singular point of (1.2). The point z = 0 is called a regular singularity if pi has a pole of order at most i for i = 1, . . . , n. Another way of characterising a regular singularity is by rewriting (1.2) with respect to the derivation θ. Multiply (1.2) with zn and use zrDr = θ(θ − 1)(θ − r + 1) to obtain an equation of the form

n n−1 θ y + q1(z)θ y + ··· + qn−1(z)θy + qn(z)y = 0. (1.3)

The condition for z = 0 to be a regular singularity comes down to qi ∈ C[[z]] for all i. Similarly we can consider a system of first order equations over C((z)), ∂y = Ay where A has now entries in C((z)). Again we call the point z = 0 regular if all entries of A are in C[[z]] and singular otherwise. We call z = 0 a regular singularity if the entries of A have a pole of order at most one. Again, when we write the system with respect to the operator θ, i.e. θy = zAy, the condition that z = 0 is a regular singularity comes down to zA having entries in C[[z]]. One also verifies that a differential equation with a regular point can be rewritten as a system with a regular point and that an equation with a regular singularity can be written as a system with a regular singularity by starting from (1.3).

Theorem 1.2.2 (Cauchy) Suppose 0 is a regular point of (1.2). Then there exist n C-linear independent Taylor series solutions f1, . . . , fn ∈ C[[z]]. More- over, any Taylor series solution of (1.2) is a C-linear combination of f1, . . . , fn. Moreover, if the coeﬃcients of (1.2) all have positive radius of convergence, the same holds for f1, . . . , fn.

This theorem is a consequence of the following statement

F.Beukers: Hypergeometric Functions, preliminary notes 8 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

d Theorem 1.2.3 (Cauchy) Consider the system of equations dz y = Ay and suppose that the entries of A are in C[[z]]. Then the system has a fundamental solution matrix Y with entries in C[[z]] and Y (0) = Id. Here Id is the n × n identity matrix. Moreover, if the entries of A have positive radius of convergence, the same holds for the entries of Y . Clearly the columns of Y form an independent set of n vector solutions of the system. Since the dimension of the solution space is at most n this means that the columns of Y form a basis of solutions in C[[z]]. There is also a converse statement. Theorem 1.2.4 Suppose that the n×n matrix A has entries in C((z)). Suppose d there is a fundamental solution matrix Y ∈ GL(n, C[[z]]) of dz Y = AY . In particular we have that Y (0) is invertible. Then z = 0 is a regular point of the system.

d −1 Proof. The proof consists of the observation that dz Y · Y has entries in C[[z]]. 2 For diﬀerential equations this theorem implies that if we have a basis of solutions i of the form fi = z (1 + O(z)), i = 0, . . . , n − 1 then z = 0 is a regular point. The extra condition on the shape of the fi is really necessary since the mere existence a basis of holomorphic solutions does not always imply that z = 0 is regular. For 2 1 2 example, the equation D y − z Dy = 0 has 1, z as basis of solutions, but z = 0 is not a regular point. Note that in the case of systemsµ the condition¶ Y (0)µ invertible is essential.¶ For z − 1 1 1 1 + z example, the system Dy = 1 y has as funda- z2 −1 1 + z 1 − z 1 mental solution matrix. If a diﬀerential equation or a system of equations with a singular point at z = 0 has a basis of solutions with components in C[[z]] we call z = 0 an apparent singularity. The proof of Cauchy’s theorems follows from the following lemma. Lemma 1.2.5 Consider the system θy = Ay where A is an n × n-matrix with entries in C[[z]]. So z = 0 is a regular singularity. Let ρ be an eigenvalue of t A(0) such that none of ρ + 1, ρ + 2,... is eigenvalue of A(0). Let (g0, . . . , gn) be an eigenvector of A(0) with eigenvalue ρ. Then the system has a solution of the ρ t form z (G1,...,Gn) with Gi ∈ C[[z]] and Gi(0) = gi for all i. Moreover, if the entries of A have positive radius of convergence, the same holds for the Gi(z).

P i Proof. Write A = i≥0 Aiz . We look for a solution y of the form y = ρ P i z i≥0 yiz , where the yi have constant entries and y0 6= 0. Substitution of y in the diﬀerential equation yields the recursion

(k + ρ)yk − A0yk = A1yk−1 + ··· + Aky0

F.Beukers: Hypergeometric Functions, preliminary notes 1.2. LOCAL THEORY 9

for k = 0, 1, 2,.... When k = 0 we see that the recursion implies that y0 is an eigenvector of A0 with eigenvalue ρ. Choose y0 to be such an eigenvector. Since ρ+k is not an eigenvalue of A0 for k = 1, 2,..., The matrix k+ρ−A0 is invertible for all k ≥ 1 and our recursion gives the yk. Now suppose that the entries of A have positive radius of convergence. This i means that there exist C, σ ∈ R>1 such that ||Ai|| ≤ Cσ . Here ||B|| denotes the norm of an n × n matrix B deﬁned by the supremum of all |Bv| as v runs over n all vectors in C of length 1. It is not hard to show that there exist k0 ∈ Z≥0 −1 −1 and λ ∈ R>0 such that ||(k + ρ − A0) || ≤ (k − λ) whenever k > k0. For future use we also see to it that k0 ≥ λ + 2C. Let M be the maximum of |yi| for i = 0, . . . , k0. Then, by using the recursion and induction on k one can show k that |yk| ≤ M(2σ) for all k ≥ 0. 2 If we have a system where z = 0 is a regular point, this means that A(0) of Lemma 1.2.5 is identically zero. Hence ρ = 0 and any non-trivial vector is an eigenvector. So we take the standard basis in Cn and obtain Cauchy’s theorems. In the following theorem we shall consider expressions of the form zA where A is a constant n × n matrix. This is short hand for X 1 zA = exp(A log z) = Ak(log z)k. k! k≥0

In particular zA is an n × n matrix of multivalued functions around z = 0. Examples are, Ã ! Ã ! 1/2 0 µ ¶ 0 1 µ ¶ z1/2 0 1 log z z 0 −1/2 = , z 0 0 = . 0 z−1/2 0 1

Theorem 1.2.6 (Fuchs) Suppose that the n × n matrix A has entries in C[[z]]. Then the system of equations θy = Ay has a fundamental matrix solution of the form S · zB, where S is an n× matrix with entries in C[[z]] and B is a constant upper triangular matrix. Any eigenvalue ρ of B is the minimum of all eigenvalues of A(0) of the form ρ, ρ + 1, ρ + 2,.... In particular, if the eigenvalues of A(0) are all distinct modulo 1, the eigenvalues of B and A(0) coincide and S(0) is invertible. Moreover, if the entries of A have positive radius of convergence, the same holds for the entries of S.

Notice that the existence of the fundamental solution matrix S · zB implies that the system is equivalent over C((z)) to θy = By, which has zB as fundamental solution matrix.

Proof. We shall prove our theorem by induction on n. When n = 1, Lemma 1.2.5 gives a solution of the form zρG(z), as desired.

F.Beukers: Hypergeometric Functions, preliminary notes 10 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

Suppose now that n > 1. Let ρ be an eigenvalue of A(0) such that none of ρ + 1, ρ + 2,... is an eigenvalue of A(0). Then there exists a solution of the form zρg where g has entries in C[[z]] and at least one of the entries has a non-zero constant term. Without loss of generality we can assume g1(0) 6= 0. Replace y by   g1(z) 0 ··· 0  g2(z) 1 ... 0   . .  y.  . .  gn(z) 0 ··· 1 Our new equation will have zρ(1, 0,..., 0)t as solution hence it has the form   ρ l2 ··· ln  0 a22 ··· a2n  θy =  . .  y  . .  0 an2 ··· ann

With l2, . . . , ln ∈ C[[z]]. According to our induction hypothesis the (n − 1) × (n − 1) system with coefficient matrix (aij)i,j=2,...,n is equivalent to a system with a constant upper triangular coefficient matrix, say C We can use this to bring our n × n-system in the form with coefficient matrix   ρ l2 ··· ln  0   .   . C  0

Now replace y1 by y1 + m2y2 + ··· + mnyn. One veriﬁes that we obtain a new ˜ ˜ system of the same form as above, except that the li have changed into l2,..., ln where ˜ ˜ (l2,..., ln) = (l2, . . . , ln) − θ(m2, . . . , mn) + (m2, . . . , mn)(ρ − C).

When ρ − r is not an eigenvalue of C we see that the equation

r 0 = (λ2, . . . , λn)z − θ(m2, . . . , mn) + (m2, . . . , mn)(ρ − C)

−1 r has (λ2, . . . , λn)(r − ρ + C) z as a solution. We can apply this principle to the terms of the power series expansion of (l2, . . . , ln). When none of the numbers ρ, ρ − 1, ρ − 2,... is an eigenvalue of C we can thus ﬁnd m2, . . . , mn ∈ C[[z]] such ˜ that the li all become zero. Our theorem is proved in this case. Suppose now that C has an eigenvalue of the form ρ − k for some k ∈ Z≥0. By our induction hypothesis there is only one such k. Using our remarks above ˜ k we can now choose m2, . . . , mn ∈ C[[z]] such that li = λiz with λi ∈ C for k i = 2, . . . , n. Now replace y1 by z y1. The top row of the coeﬃcient matrix of the

F.Beukers: Hypergeometric Functions, preliminary notes 1.2. LOCAL THEORY 11

new equation now reads (ρ − k, λ2, . . . , λn), while the other rows stay the same. Consequently the coeﬃcient matrix now contains only elements from C and it is upper triangular. This proves our theorem. 2 The converse of Theorem 1.2.6 need not hold, so a system with a fundamental B matrix solution of the form S · z need not have a regular singularityµ at z =¶ 0. 1 1 + z Consider for example the fundamental solution matrix Y = . 1 − z 1 However, if the system comes directly from a diﬀerential equation we do have a positive statement.

Theorem 1.2.7 Suppose the diﬀerential equation (1.2) has a basis of solutions of the form B (g1(z), . . . , gn(z))z , g1(z), . . . , gn(z) ∈ C[[z]] where B is a constant matrix. Then (1.2) has a regular singularity at z = 0.

Proof. Under construction.

Corollary 1.2.8 Suppose that the coeﬃcients of (1.2) converge in a region D = {z|0 < |z| < σ}. Suppose that in every sector of D with 0 as vertex we have a λ basis of solutions f1, . . . , fn of (1.2) and λ ∈ R such that z fi(z) tends to 0 as z → 0. Then z = 0 is a regular singularity of (1.2)

The condition that in every sector the solutions are polynomially bounded by 1/|z| is called the condition of moderate growth.

Proof. Choose a sector S in D with a basis of solutions f1, . . . , fn. Let γ be a simple closed path which goes around zero once. Now continue f1, . . . , fn ana- ˜ ˜ lytically along γ until we return to our sector S. The continuations f1,..., fn are still solutions of (1.2). Hence there exists a constant matrix M such that ˜ ˜ (f1,..., fn) = (f1, . . . , fn)M in S. We call M the monodromy matrix.√ corre- 2π −1B sponding to the fi and γ. Choose a constant matrix B such that e = M. −B Then the n-tuple of functions (f1, . . . , fn)z has trivial monodromy around z = 0, and hence these functions can be continued to the punctured disc D. −B The moderate growth condition now implies that the entries of (f1, . . . , fn)z are in fact meromorphic functions. We can now apply our previous theorem. 2 Let A be as in Theorem 1.2.6. The eigenvalues of A(0) are called the local exponents at z = 0 of the system. If we have a differential equation where z = 0 is a regular singularity, we first write it in the form (1.3) and then as a system. One verifies that the local exponents of the system are the solutions of n n−1 the equation x +q1(0)x +···+qn−1(0)x+qn(0) = 0. We call this equation the indicial equation and its solutions the local exponents of the equation at z = 0. 2 Note that if we choose a different local parameter t via t = c1z+c2z +··· , c1 6= 0

F.Beukers: Hypergeometric Functions, preliminary notes 12 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS and rewrite our equation or system with respect to t, then the local exponents at t = 0 are the same as the original exponents. This is worked out in the following exercise.

Exercise 1.2.9 Show that there exists a powerseries g(t) in t with constant coeﬃcient 1, such that Dz = g(t)Dt. Now show by induction on n that for every n there exist powerseries g1, . . . , gn−1 with vanishing constant term such n n n n−1 that Dz = g(t) Dt + g1(t)Dt + ··· + gn−1(t)Dt. Show that q˜i(0) = qi(0) for i = 1, . . . , n.

Remark 1.2.10 Notice that if we replace y by zµw, the diﬀerential equation for w reads

n n−1 (D + µ) w + q1(z)(D + µ) w + ··· + qn−1(z)(D + µ)w + qn(z)w = 0. In particular, the local exponents have all decreased by µ.

Exercise 1.2.11 Show that the local exponents at a regular point read 0, 1, . . . , n− 1.

Exercise 1.2.12 Consider the linear diﬀerential equation

(z3 + 11z2 − z)y00 + (3z2 + 22z − 1)y0 + (z + 3)y = 0

. Show that the local exponents at z = 0 are 0, 0 and determine the recursion relation for the holomorphic solution near z = 0. Determine also the ﬁrst few terms of the expansions of a basis of solutions near z = 0.

1.3 Fuchsian equations

In this section our differential field will be C(z), the field of rational functions in z and we shall consider our differential equations and n × n-systems over this field. Consider the linear differential equation

(n) (n−1) 0 y + p1(z)y + ··· + pn−1(z)y + pn(z)y = 0, pi(z) ∈ C(z) (1.4) To study this diﬀerential equation near any point P ∈ P1 we choose a local parameter t ∈ C(z) at this point (usually t = z − P if P ∈ C and t = 1/z if P = ∞), and rewrite the equation with respect to the new variable t. We call the point P a regular point or a regular singularity if this is so for the equation in t at t = 0. It is not diﬃcult to verify that a point P ∈ C is regular if and only if the i pi have no pole at P . It is a regular singularity if and only if limz→P (z − P ) pi(z) exists for i = 1, . . . , n. The point ∞ is regular or a regular singularity if and only i if limz→∞ z pi(z) exists for i = 1, . . . , n.

F.Beukers: Hypergeometric Functions, preliminary notes 1.3. FUCHSIAN EQUATIONS 13

Definition 1.3.1 A differential equation over C(z) or a system of first order equations over C(z) is called Fuchsian if all points on P1 are regular or a regular singularity.

The form of Fuchsian systems is particularly simple. Let our n × n-system be given by Dy = Ay where the entries of A are in C(z). Let S = {p1, . . . , pr} be the set of ﬁnite singular points. If we have a Fuchsian system of equations then there exist constant matrices A1,...,Ar such that A A A(z) = 1 + ··· + r . z − p1 z − pr Pr The point ∞ is regular if and only if i=1 Ai = 0. Let P ∈ P1 be any point which is regular or a regular singularity. Let t be a local parameter around this point and rewrite the equation (1.4) with respect to the variable t. The corresponding indicial equation will be called the indicial equation of (1.4) at P . The roots of the indicial equation at P are called the local exponents of (1.4) at P . This procedure can be cumbersome and as a shortcut we use the following lemma to compute indicial equations.

Lemma 1.3.2 Let P ∈ C be a regular point or regular singularity of (1.4). Let i ai = limz→P (z − P ) pi(z) for i = 1, . . . , n. The indicial equation at P is given by

X(X − 1) ··· (X − n + 1) + a1X(X − 1) ··· (X − n + 2) + ··· + an−1X + an = 0.

i When ∞ is regular or a regular singularity, let ai = limz→∞ z pi(z) for i = 1, . . . , n. The indicial equation at ∞ is given by

X(X + 1) ··· (X + n − 1) − a1X(X + 1) ··· (X + n − 2) + ··· n−1 n +(−1) an−1X + (−1) an = 0.

Proof. Exercise

Theorem 1.3.3 (Fuchs’ relation) Suppose (1.4) is a Fuchsian equation. Let 1 ρ1(P ), . . . , ρn(P ) the set of local exponents at any P ∈ P . Then, µ ¶ µ ¶ X n n (ρ (P ) + ··· + ρ (P ) − ) = −2 1 n 2 2 P ∈P1 Since the local exponents at a regular point are always 0, 1, . . . , n − 1 the terms in the summation are zero when P is a regular point. So, in fact, the summation in this theorem is a ﬁnite sum.

F.Beukers: Hypergeometric Functions, preliminary notes 14 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

Proof. From the explicit shape of the indicial equations, given in the Lemma above, we infer that for P ∈ C, µ ¶ n ρ (P ) + ··· + ρ (P ) = − res (p (z)dz) 1 n 2 P 1 and µ ¶ n ρ (∞) + ··· + ρ (∞) = − − res (p (z)dz). 1 n 2 ∞ 1 ¡ ¢ Substract n on both sides and add over all P ∈ P1. Using the fact that P 2 P ∈P1 resP (p1(z)dz) = 0 yields our theorem. 2

Exercise 1.3.4 Let a, b, c ∈ C. Determine all singularities and their local exponents of the so-called hypergeometric diﬀerential equation

z(z − 1)F 00 + ((a + b + 1)z − c)F 0 + abF = 0.

For non-integral c write the recurrence relation for the coeﬃcients of the power series expansions of the solutions around z = 0.

From Cauchy’s theorem of the previous section follows automatically Theorem 1.3.5 (Cauchy) Suppose P ∈ C is a regular point of the system of equations Dy = Ay. Then there exist n C-linear independent vector solutions y1,..., yn with Taylor series entries in z − P with positive radius of convergence. Moreover, any Taylor series solution of the system is a C-linear combination of y1,..., yn.

Corollary 1.3.6 Any analytic solution of Dy = Ay near a regular point can be continued analytically along any path in C not meeting any singularity.

1 Let S be the set of singularities of Dy = Ay and let z0 ∈ P \ S. Let vy1,..., yn be an independent set of analytic solutions around z0. They are the columns 1 of the fundamental solution matrix Y . Let γ ∈ π1(P \ S, z0). After analytic continuation of Y along γ we obtain a new fundamental solution matrix Y˜ . Hence there exists a square matrix M(γ) ∈ GL(n, C) such that Y˜ = Y · M. The map 1 ρ : π1(P \ S) → GL(n, C) given by ρ : γ 7→ M(γ) is a group homomorphism and its image is called the monodromy group of the system.

1.4 Riemann-Hilbert correspondence

Suppose we are given an n×n Fuchsian system of ﬁrst order equations, Dy = Ay, where A has entries in C(z). Let S ⊂ P1 be the set of singular points and write S = {s1, s2, . . . , sm}. Without loss of generality we can assume that ∞ 6∈ S.

F.Beukers: Hypergeometric Functions, preliminary notes 1.4. RIEMANN-HILBERT CORRESPONDENCE 15

Fix a base point z0 6∈ S and let γi be a simple closed beginning and ending in z0 and which contains only the point si as singularity in its interior. Choose a fundamental solution matrix of the system near z0. Corresponding to this choice we can associate to each loop γi the monodromy matrix Mi. The monodromy representation is determined by these matrices. If we order the si such that 1 γ1 ··· γm = 1 in π1(P \ S, z0) we have moreover that M1 ··· Mm = Id. We have the following question.

Question 1.4.1 (Riemann-Hilbert problem) Suppose S = {s1, . . . , sm} is a 1 ﬁnite subset of P . Again we assign a simple loop γi to each si and order them 1 in such a way that γ1 ··· γm = 1 in π1(P \ S, z0). To each si ∈ S we assign a matrix Mi ∈ GL(n, C) such that M1 ··· Mm = Id. Does there exist a Fuchsian system with singularities only in S whose monodromy representation ρ is (up to conjugation) given by ρ : γi 7→ Mi for i = 1, . . . , n ?

The answer is as follows,

Theorem 1.4.2 (Plemelj, 1906) Let notations be as in the above problem. Then there exists an n×n Fuchsian system Dy = Ay whose monodromy representation is given by the matrices Mi. Moreover, the system can be chosen in such a way that the singular set is given by S ∪ {a}, where a is an arbitrary point in P1 \ S and forms an apparent singularity of the system.

So we see that the answer to our question is almost aﬃrmative. It may be necessary to have an extra singularity in the Fuchsian system. It was recently shown by Bolibruch that there are examples of representations of the fundamental 1 group π1(P \ S) where any corresponding Fuchsian system requires an extra singularity. It is also known that we do not need an extra singularity if the representation is irreducible or if one of the matrices Mi is semi-simple. To prove Plemelj’s theorem we shall use the following theorem.

Theorem 1.4.3 (Birkhoﬀ, Grothendieck) Any holomorphic vector bundle on 1 P of rank n is of the form O(m1) ⊕ · · · ⊕ O(mn) where the mi are integers and where O(m) is the line bundle on P1 corresponding to the divisor m∞.

More particularly we shall need the following Corollary which says that any rank n holomorphic vector bundle over P1 has n meromorphic sections which form a basis of the ﬁber above every point in C.

Corollary 1.4.4 We cover P1 with C and an open disc U around ∞. For any 1 holomorphic vector bundle E over P there exist integers m1, . . . , mn and a local n trivialisation f : C×C → E with the property that for any local trivialisation hU : n −1 ∗ n m1 mn U×C → E the map hU ◦f of U ×C to itself has the form Z·diag(z , . . . , z ), where Z : U → GL(n, C) is holomorphic.

F.Beukers: Hypergeometric Functions, preliminary notes 16 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

Proof of Plemelj’s theorem. Without loss of generality we can assume that a = ∞. Choose z0 in such a way that it does not lie on any line connecting two points of S. For each i we now draw the line connecting z0 and si and denote the half line connecting starting at si and not containing z0 by li. Denote the open k set C \ (∪i=1li) by U0. Now choose ² > 0. To each i we associate the open set Di ⊂ C given by all points z ∈ C whose distance to li is less than ². By choosing ² sufficiently small we can see to it that the Di are disjoint. Choose R such that |si| < R for all i and denote by U∞ the set z : |z| > R together with the point 1 at infinity. So the open sets U0,U∞,D1,...,Dk form an open cover of P . Using this cover we construct a vector bundle of rank √r. 2π −1Gi For each i we choose a matrix Gi such that e = Mi. Note there is some ambiguity in the choice of Gi. We define the holomorphic map fi : Di ∩ U0 → Gi GL(n, C) by some choice of (z − si) . Note that U0 ∩ U∞ consists of m open sectors around the point ∞. We call these sectors T1,...,Tm and order them so that Ti lies between li and li+1 (indices considered modulo m + 1). We define the map h : U0 ∩ U∞ → GL(n, C) by h(z) = M1 ··· Mi if z ∈ Ti. We define the holomorphic map gi : Di ∩ U∞ → GL(n, C) by the function which coincides with M1 ··· Mifi on Ti. Note that by analytic continuation this implies that gi coincides with M1 ··· Mi−1fi on Ti−1. As a consequence we have for each i that gi = hfi on U0 ∩ U∞ ∩ Di. n n Construct a vector bundle as follows. Glue U0 ×C to Di ×C via the equivalence relation −1 (z, v) ∼ (zi, vi) ⇐⇒ z = zi ∈ U0 ∩ Di and vi = fi v. n n Glue U∞ × C to Di × C via the equivalence relation −1 (z, v) ∼ (zi, vi) ⇐⇒ z = zi ∈ U∞ ∩ Di and vi = gi v. n n Finally glue U0 × C to U∞ × C via the relation

(z0, v0) ∼ (z∞, v∞) ⇐⇒ z = zi ∈ U0 ∩ U∞ and v∞ = hv0.

Because of the relation gi = hfi for all i this can be done in a compatible way. We thus obtain a holomorphic vector bundle E over P1. According to the Corollary of the Birkhoﬀ-Grothendieck theorem there is a meromorphic local trivialisation n f : C × C → E. This implies that there exist holomorphic functions t0 : U0 → GL(n, C) and ti : Di → GL(n, C) and t∞ : U∞ → GL(n, C) with the −1 −1 −1 G∞ property that t0 ti = fi for i = 1, . . . , m and t0 t∞z = h, where G∞ = diag(m1, . . . , mn). Now observe that t0 is a fundamental solution matrix of the system of equations −1 Dy = Ay where A = (Dt0)t0 . Note that A has holomorphic entries on U0. The continuation of t0 to Di is given by tifi. Hence the continuation of A to Di has the form

t G t−1 Gi −Gi −1 −1 i i i (D(t∞(z − si) )(z − si) ti = (Dti)ti + . z − si

F.Beukers: Hypergeometric Functions, preliminary notes 1.5. FUCHSIAN EQUATIONS OF ORDER TWO 17

Similarly the continuation of A to U∞ has the form

t G t−1 (D(t zG∞ h−1)hz−G∞ t−1 = (Dt )t−1 + ∞ ∞ ∞ . ∞ ∞ ∞ ∞ z Gi Note that limz→∞ A = 0. Hence our system is Fuchsian. Since t0 = ti(z − si) in every Di, the functions t0 have the correct local monodromy behaviour at every point si. We have thus found our desired Fuchsian system of equations. 2

1.5 Fuchsian equations of order two

It is an interesting exercise to write down all Fuchsian diﬀerential equations with a given number of singular points. Let us start with ﬁrst order Fuchsian equations Exercise 1.5.1 Show that any Fuchsian equation of order one can be written in the form µ ¶ dy A A + 1 + ... + k y = 0 dz z − a1 z − ak for suitable ai,Ai ∈ C. Solve this equation. Let us now turn to higher order Fuchsian equations Exercise 1.5.2 Show that any Fuchsian equation having only ∞ as singular dny point is of the form dzn = 0. More generally, Fuchsian equations having only one singularity are not very interesting since, by a fractional linear transformation, the singularity can be moved to ∞. Exercise 1.5.3 Show that any Fuchsian equation having only 0 and ∞ as singular points is of the form

n (n) n−1 (n−1) 0 z y + a1z y + ··· + an−1zy + any = 0 for suitable a1, . . . , an ∈ C. Verify that the indicial equation has the form

X(X − 1) ··· (X − n + 1) + a1X ··· (X − n + 2) + ··· + an−1X + an = 0. Equations such as these are known as Euler equations. Suppose that the local exponents at z = 0 are all distinct. Then write down a basis of solutions. More generally, any Fuchsian equation with two singularities can be transformed into an Euler equation. The underlying reason why Fuchsian equations with one or two singularities are not very exciting is that the fundamental groups of P1 \ ∞ and P1 \{0, ∞} are trivial and Z respectively, i.e. they are both abelian groups. Interesting equations can be expected when there are three or more singular points.

F.Beukers: Hypergeometric Functions, preliminary notes 18 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

Exercise 1.5.4 Suppose we have a second order Fuchsian equation with singularities 0, 1, ∞ and suppose the local exponents at these points are given by the following scheme, 0 1 ∞ 0 0 a 1 − c c − a − b b The second exponent at 1 is chosen to satisfy Fuchs’s relation for exponents. Show that the corresponding second order equation is uniquely determined and reads,

z(z − 1)F 00 + ((a + b + 1)z − c)F 0 + abF = 0.

This is the hypergeometric equation with parameters a, b, c.

Suppose we have a second order equation with three singularities, say A, B, C. To eqach singularity we have local exponents which we put in the following (Rie- mann) scheme,

ABC α β γ α0 β0 γ0 Via a M¨obiustransformation we can map A, B, C to any three distinct points of P1. Let us take the mapping A, B, C → 0, 1, ∞. So we have to deal with the Fuchsian equation having Riemann scheme

0 1 ∞ α β γ α0 β0 γ0

If we multiply the solutions of the latter equation by zµ we obtain a set of functions that satisfy the Fuchsian equation with Riemann scheme

0 1 ∞ α + µ β γ − µ α0 + µ β0 γ0 − µ

A fortiori, after multiplication of the solutions with z−α0 (1 − z)−β0 we obtain a Fuchsian equation with a scheme of the form

0 1 ∞ α00 β00 γ00 0 0 1 − α00 − β00 − γ00

Hence any second order Fuchsian equation with three singularities can be transformed into a hypergeometric equation. Any hypergeometric equation is uniquely

F.Beukers: Hypergeometric Functions, preliminary notes 1.5. FUCHSIAN EQUATIONS OF ORDER TWO 19 determined by its local exponents and, a fortiori, any second order Fuchsian equation is uniquely determined by the location of its singularities and their local exponents. Here are some remarks on the solutions of the hypergeometric equation. When c is not integral a basis of solutions is given by

X (a) (b) F (a, b, c|z) := n n zn. (1.5) (c)nn! and z1−cF (a + 1 − c, b + 1 − c, 2 − c|z)

The Pochhammer symbol (x)n is deﬁned by (x)0 = 1 and (x)n = x(x + 1) ··· (x + n − 1). The function F (a, b, c|z) is known as Gauss’ hypergeometric function.

Exercise 1.5.5 Show directly that the power series (1.5) satisﬁes the diﬀerential equation d z(D + a)(D + b)F = D(D + c − 1)F,D = z dz Using the above theory it is very simple to prove some quadratic relations between hypergeometric functions, such as

F (a, b, a + b + 1/2|4t − 4t2) = F (2a, 2b, a + b + 1/2|t) and F (a, b, a + b + 1/2|t2/(4t − 4)) = (1 − t)aF (2a, a + b, 2a + 2b|t). they were discovered by E.Kummer. Let us prove for example the quadratic relation

F (a, b, a + b + 1/2|t2/(4t − 4)) = (1 − t)aF (2a, a + b, 2a + 2b|t). Substitute z = t2/(4t−4) in the hypergeometric equation with parameters a, b, a+ b+1/2. We obtain a new Fuchsian equation. The map t → z = t2/(4t−4) ramiﬁes above 0, 1 in t = 0, 2 respectively. Above z = 1 we have the point t = 2, above z = 0 the point t = 0 and above z = ∞ the two points t = 1, ∞. Notice that our equation has local exponents 0, 1/2 in z = 1. Hence the new equation has local exponents 0, 1 in t = 2, with regular solutions, and t = 2 turns out to be a regular point. At t = 0 we get the local exponents 0, 2(1/2 − a − b) and in t = 1, ∞, the points above z = ∞, we have the local exponents a, b and a, b. Thus our equation in t has again three singular points and Riemann scheme

0 1 ∞ 0 a a 1 − 2a − 2b b b

F.Beukers: Hypergeometric Functions, preliminary notes 20 CHAPTER 1. ORDINARY LINEAR DIFFERENTIAL EQUATIONS

By the method sketched above, one easily sees that (1−t)aF (2a, a+b, 2a+2b|t) is a solution of this equation. Moreover, this is the unique (up to a constant factor) solution holomorphic near t = 0. At the same time F (a, b, a + b + 1/2|t2/(4t − 4)) is a solution, and by the uniquess equality follows.

Exercise 1.5.6 Prove in a similar way the equality

F (a, b, a + b + 1/2|4z − 4z2) = F (2a, 2b, a + b + 1/2|z).

F.Beukers: Hypergeometric Functions, preliminary notes Chapter 2

Gauss hypergeometric functions

2.1 Deﬁnition, ﬁrst properties

Let a, b, c ∈ R and c 6∈ Z≤0. Deﬁne Gauss’ hypergeometric function by X (a) (b) F (a, b, c|z) = n n zn. (2.1) (c)nn!

The Pochhammer symbol (x)n is deﬁned by (x)0 = 1 and (x)n = x(x + 1) ··· (x + n − 1). The radius of convergence of (2.1) is 1 unless a or b is a non-positive integer, in which cases we have a polynomial.

Examples.

Here K(z) is the Jacobi’s elliptic integral of the ﬁrst kind given by Z 1 dx K(z) = p . 2 2 2 0 (1 − x )(1 − z x )

n 2 n The polynomials Pn,Tn given by Pn = (1/n!)(d/dz) (1 − z ) and Tn(cos z) = cos(nz) are known as the Legendre and Chebyshev polynomials respectively. They are examples of orthogonal polynomials.

21 22 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

One easily verifies that (2.1) satisfies the linear differential equation d z(D + a)(D + b)F = D(D + c − 1)F,D = z . dz Written more explicitly, z(z − 1)F 00 + ((a + b + 1)z − c)F 0 + abF = 0. (2.2) There exist various ways to study the analytic continuation of (2.1), via Eu- ler integrals, Kummer’s solutions and Riemann’s approach. The latter will be discussed in later sections. The Euler integral reads Z Γ(c) 1 F (a, b, c|z) = tb−1(1 − t)c−b−1(1 − tz)−adt (c > b > 0) Γ(b)Γ(c − b) 0 and allows choices of z with |z| > 1. The restriction c > b > 0 is included to ensure convergence of the integral at 0 and 1. We can drop this condition if we take the Pochhammer contour γ given by

0 1 X

as integration path. Notice that the integrand acquires the same value after analytic continuation along γ. It is a straightforward exercise to show that for any b, c − b 6∈ Z we have Z Γ(c) 1 b−1 c−b−1 −a F (a, b, c|z) = 2πib 2πi(c−b) t (1 − t) (1 − tz) dt Γ(b)Γ(c − b) (1 − e )(1 − e ) γ Kummer gave the following 24 solutions to (4.1) F (a, b, c|z) = (1 − z)c−a−bF (c − a, c − b, c|z) = (1 − z)−aF (a, c − b, c|z/(z − 1)) = (1 − z)−bF (a − c, b, c|z/(z − 1))

z1−cF (a − c + 1, b − c + 1, 2 − c|z) = z1−c(1 − z)c−a−bF (1 − a, 1 − b, 2 − c|z) = z1−c(1 − z)c−a−1F (a − c + 1, 1 − b, 2 − c|z/(z − 1)) = z1−c(1 − z)c−b−1F (1 − a, b − c + 1, 2 − c|z/(z − 1))

F.Beukers: Hypergeometric Functions, preliminary notes 2.1. DEFINITION, FIRST PROPERTIES 23

F (a, b, a + b − c + 1|1 − z) = x1−cF (a − c + 1, b − c + 1, a + b − c + 1|1 − z) = z−aF (a, a − c + 1, a + b − c + 1|1 − 1/z) = z−bF (b − c + 1, b, a + b − c + 1|1 − 1/z)

(1 − z)c−a−bF (c − a, c − b, c − a − b + 1|1 − z) = (1 − z)c−a−bz1−cF (1 − a, 1 − b, c − a − b + 1|1 − z) = (1 − z)c−a−bza−cF (1 − a, c − a, c − a − b + 1|1 − 1/z) = (1 − z)c−a−bzb−cF (c − b, 1 − b, c − a − b + 1|1 − 1/z)

z−aF (a, a − c + 1, a − b + 1|1/z) = z−a(1 − 1/z)c−a−bF (1 − b, c − b, a − b + 1|1/z) = z−a(1 − 1/z)c−a−1F (a − c + 1, 1 − b, 2 − c|1/(1 − z)) = z−a(1 − 1/z)−aF (a, c − b, a − b + 1|1/(1 − z))

z−bF (b, b − c + 1, b − a + 1|1/z) = z−b(1 − 1/z)c−a−bF (1 − a, c − a, b − a + 1|1/z) = z−b(1 − 1/z)c−b−1F (b − c + 1, 1 − a, 2 − c|1/(1 − z)) = z−b(1 − 1/z)−bF (b, c − a, b − a + 1|1/(1 − z))

Strictly speaking, the above six 4-tuples of functions are only distinct when c, c− a − b, a − b 6∈ Z. If one of these numbers is an integer we find that there are other solutions containing logarithms. For example, when c = 1 we find that z1−c becomes log z and a second solution near z = 0 reads " µ ¶# X∞ (a) (b ) Xn 1 1 2 (log z)F (a, b, 1|z) + n n zn + − . (n!)2 a + k − 1 b + k − 1 k n=1 k=1 Notice that this solution can be obtained by taking the difference of solutions z1−cF (a − c + 1, b − c + 1, 2 − c|z) − F (a, b, c|z), divide it by c − 1 and take the limit as c → 1. Later it will turn out that Riemann’s approach to hypergeometric functions gives a remarkably transparent insight into these formulas as well as the quadratic transformations of Kummer and Goursat. Examples of such transformations are

F (a, b, a + b + 1/2|4z − 4z2) = F (2a, 2b, a + b + 1/2|z) and

F (a, b, a + b + 1/2|z2/(4z − 4)) = (1 − z)aF (2a, a + b, 2a + 2b|z).

F.Beukers: Hypergeometric Functions, preliminary notes 24 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

Finally we mention the 6 contiguous functions

F (a ± 1, b, c|z),F (a, b ± 1, c|z),F (a, b, c ± 1|z).

Gauss found that F (a, b, c|z) and any two contiguous functions satisfy a linear relation with coeﬃcients which are linear polynomials in z or constants, for example,

(c − a)F (a − 1, b, c|z) + (2a − c − az + bz)F (a, b, c|z) + a(z − 1)F (a + 1, b, c|z) = 0.

Notice also that F 0(a, b, c|z) = (ab/c)F (a + 1, b + 1, c + 1|z). These observations are part of the following theorem.

Theorem 2.1.1 Suppose a, b 6≡ 0, c(mod Z) and c 6∈ Z. Then any function F (a + k, b + l, c + m|z) with k, l, m ∈ Z equals a linear combination of F,F 0 with rational functions as coeﬃcients.

c d F (a, b, c + 1|z) = (z(1 − z) + c − a − b)F (a, b, c|z) (c − a)(c − b) dz 1 d F (a, b, c − 1|z) = (z + c − 1)F (a, b, c|z) c − 1 dz

d Hence there exists a linear differential operator Lk,l,m ∈ C(z)[ dz ] such that F (a + k, b + l, c + m|z) = Lk,l,mF (a, b, c|z). Since F satifies a second order linear differential equation, Lk,l,mF can be written as a C(z)-linear combination of F and F 0. 2 In general we shall call any function F (a + k, b + l, c + m|z) with k, l, m ∈ Z contiguous with F (a, b, c|z). Thus we see that, under the assumptions of Theorem 2.1.1, any three contiguous functions satisfy a C(z)-linear relation. For many more identities and formulas we refer to [AS] and [E].

2.2 Monodromy of the hypergeometric function

Let us now turn to the monodromy of the hypergeometric equation. Consider the three loops g0, g1, g∞ which satisfy the relation g0g1g∞ = 1.

F.Beukers: Hypergeometric Functions, preliminary notes 2.2. MONODROMY OF THE HYPERGEOMETRIC FUNCTION 25

0 1 X

We denote the corresponding monodromy matrices by M0,M1,M∞. They also satisfy M0M1M∞ = 1 and M0,M∞ generate the monodromy group. Since the local exponents at 0, 1, ∞ are 0, 1 − c, 0, c − a − b and a, b respectively, the eigenvalues of the matrices M0,M1 and M∞ are 1, exp(2πi(1 − c)), 1, exp(2πi(c − a − b)) and exp(2πia), exp(2πib) respectively. The monodromy group can be −1 considered as being generated by M0,M∞ and we know that M∞M0 = M1 has eigenvalue 1. This scant information already suﬃces to draw some important conclusions. Lemma 2.2.1 Let A, B ∈ GL(2, C). Suppose that AB−1 has eigenvalue 1. Then there exists a common eigenvector of A, B if and only if A, B have a common eigenvalue.

Proof. Notice that ker(A − B) has dimension at least 1. If the dimension were 2 we would have A = B and our lemma would be trivial. So we can assume dim(ker(A − B)) = 1. In this proof we let v ∈ ker(A − B), v 6= 0. Suppose there exists a common eigenvector, w say, of A, B with eigenvalues λA, λB. If these eigenvalues are equal, we are done. Suppose they are not equal. Then w, v span C2. Choose α, β such that Av = αv + βw. Since Av = Bv we also have Bv = αv + βw. Hence with respect to the basis v, w the matrices of A, B read µ ¶ µ ¶ α β α β 0 λA 0 λB Hence they have the common eigenvalue α. Suppose A, B have a common eigenvalue λ. If v is an eigenvectore of A we are done, since Av = Bv implies that it is also an eigenvector of B. So suppose v is not an eigenvector of A. Consider the vector w = (A − λ)v. Since A − λ has 2 non-trivial kernel we have < w >C= (A − λ)C . In particular, (A − λ)w is a scalar multiple of w, i.e. w is an eigenvector of A. We also have w = (B − λ)v and a similar argument shows that w is an eigenvector of B. Hence A, B have a common eigenvector. 2

F.Beukers: Hypergeometric Functions, preliminary notes 26 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

Corollary 2.2.2 The monodromy group of (4.1) acts reducibly on the space of solutions if and only if at least one of the numbers a, b, c − a, c − b is integral.

Proof. This follows by application of the previous lemma to the case A = −1 −1 −1 M∞,B = M0 . Since M1 = M∞M0 the condition that AB has eigenvalue 1 is fullﬁlled. Knowing the eigenvalues of M0,M∞ one easily checks that equality of eigenvalues comes down to the non-empty intersection of the sets {0, c} and {a, b} considered modulo Z.

Deﬁnition 2.2.3 A hypergeometric equation is called reducible if its monodromy group is reducible. A hypergeometric equation is called abelian if its monodromy group is abelian.

Typical examples of abelian equations are (4.1) with a = c = 0 having solutions 1, (1 − z)−(b+1) and a = b = 1, c = 2 having solutions 1/z, log(1 − z)/z. Here is a simple necessary condition for abelian equations, which has the pleasant property that it depends only on a, b, c(mod Z).

Lemma 2.2.4 If (4.1) is abelian then at least two of the numbers a, b, c − a, c − b are integral.

Proof. Abelian monodromy implies reducibility of the monodromy, hence at least one of the four numbers is integral. Let us say a ∈ Z, the other cases can be dealt with similarly. It suffices to show that in at least one of the points 0, 1, ∞ the local exponent difference of (4.1) is integral. Then clearly, 1 − c ∈ Z implies c − a ∈ Z, c − a − b ∈ Z implies c − b ∈ Z and a − b ∈ Z implies b ∈ Z. Suppose that all local exponent differences are non-integral. In particular the eigenvalues of each of the generating monodromy elements M0,M1,M∞ are distinct. Then abelian monodromy implies that the monodromy group acts on the solution space in a completely reducible way as a sum of two one-dimensional representations. In particular the generators of these representations are functions of the form zλ(1 − z)µq(z) zλ0 (1 − z)µ0 p(z) where p(z), q(z) are polynomials with the property that they do not vanish at z = 0 or 1. The local exponents can be read off immediately, λ, λ0 at 0, µ, µ0 at 1 and −λ − µ − deg(q), −λ0 − µ0 − deg(p) at ∞. The sum of the local exponents must be 1, hence − deg(p) − deg(q) = 1. Clearly this is a contradiction. 2

Lemma 2.2.5 Suppose that A, B ∈ GL(2, C) have disjoint sets of eigenvalues −1 2 and suppose that AB has eigenvalue 1. Then, letting X + a1X + a2 and 2 X + b1X + b2 be the characteristic polynomials of A, B, we have up to common conjugation, µ ¶ µ ¶ 0 −a 0 −b A = 2 ,B = 2 . 1 −a1 1 −b1

F.Beukers: Hypergeometric Functions, preliminary notes 2.2. MONODROMY OF THE HYPERGEOMETRIC FUNCTION 27

Proof. Choose v ∈ ker(A − B) and w = Av = Bv. Since A, B have disjoint eigenvalue sets, v is not an eigenvector of A and B. Hence w, v form a basis of C2. With respect to this basis A, B automatically obtain the form given in our Lemma. 2

Corollary 2.2.6 Suppose that (4.1) is irreducible. Then, up to conjugation, the monodromy group depends only on the values of a, b, c modulo Z.

Let us now assume that a, b, c ∈ R, which is the case most frequently studied. The eigenvalues of M0,M1,M∞ then lie on the unit circle.

Deﬁnition 2.2.7 Let R,S be two disjoint ﬁnite subsets of the unit circle of equal cardinality. The sets R,S are said to interlace if every segment on the unit circle, connecting two points of R, contains a point of S.

Lemma 2.2.8 Let A, B be non-commuting elements of GL(2, C). Suppose that the eigenvalues of A, B have absolute value 1 and that AB−1 has eigenvalue 1. Let G be the group generated by A, B. Then there exists a unique (up to a constant factor) non-trivial hermitian form F on C2 such that F (g(x), g(y)) = F (x, y) for every g ∈ G and every pair x, y ∈ C2. Moreover,

F degenerate ⇐⇒ A, B have common eigenvalues

Supposing A, B have disjoint eigenvalue sets, we have in addition,

F deﬁnite ⇐⇒ eigenvalues of A, B interlace

F indeﬁnite ⇐⇒ eigenvalues of A, B do not interlace We call these three cases the euclidean, spherical and hyperbolic case respectively.

Proof. Let v ∈ ker(A − B) and w = Av. Suppose ﬁrst that v, w form a basis of C2. Of course, with respect to this basis A and B have the form given in the previous lemma. In particular we see that A, B cannot have the same characteristic equation, since this would imply that A = B. We have to ﬁnd a hermitean form F such that

F (gv, gv) = F (v, v) F (gv, gw) = F (v, w) F (gw, gv) = F (w, v),F (gw, gw) = F (w, w)

2 2 for every g ∈ G. It suﬃces to take g = A, B. Let X +a1X +a2 and X +b1X +b2 be the characteristic polynomials of A, B. Since the roots are on the unit circle we have a2a¯2 = 1, a2a¯1 = a1 and similarly for b1, b2. Let us ﬁrst take g = A. Then F (Av, Av) = F (v, v) implies

F (w, w) = F (v, v).

F.Beukers: Hypergeometric Functions, preliminary notes 28 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

The conditions F (Av, Aw) = F (v, w) and F (Aw, Av) = F (w, v) imply F (w, A2v) = 2 2 F (v, w) and F (A v, w) = F (w, v). Hence, using A = −a1A − a2,

−a¯1F (w, w) − a¯2F (w, v) = F (v, w) (2.3)

−a1F (w, w) − a2F (v, w) = F (w, v) (2.4) −1 Because of the relations a2 =a ¯2 and a2a¯1 = a1 these equations are actually the same. The condition F (Aw, Aw) = F (w, w) yields F (A2v, A2v) = F (w, w) and hence 2 2 |a1| F (w, w) + a1a¯2F (w, v) +a ¯1a2F (v, w) + |a2| F (v, v) = F (w, w). 2 Using |a2| = 1, a2a¯1 = a1 and F (w, w) = F (v, v) this is equivalent to

a1a¯1F (w, w) + a1a¯2F (w, v) + a1F (v, w) = 0 which is precisely (2.3) times a1. Hence A-invariance of F is equivalent to

F (v, v) = F (w, w),F (w, v) + a1F (w, w) + a2F (v, w) = 0. Invariance of F with respect to B yields the additional condition

F (w, v) + b1F (w, w) + b2F (v, w) = 0. Since A and B do not have the same characteristic equation the solutionspace for F is one-dimensional. When a2 = b2 a solution is given by 1/2 −1/2 F (w, w) = F (v, v) = 0,F (w, v) = (−a2) ,F (v, w) = (−a2) , when a2 6= b2 a solution is given by a − b F (w, w) = F (v, v) = 1,F (w, v) = ², F (v, w) =², ¯ ² = 1 1 . b2 − a2

We formally take ² = ∞ if a2 = b2. In both cases cases we see that F is definite, degenerate, indefinite according to the conditions |²| < 1, |²| = 1, |²| > 1. It now a straightforward excercise to see that these inequalities correspond to interlacing, coinciding or non-interlacing of the eigenvalues of A and B. We are left with the case when v is an eigenvector of A and B. Let α be the eigenvalue. If both A and B have only eigenvalues α they automatically commute, which case is excluded. So either A or B has an eigenvalue different from α. Let us say that A has the distinct eigenvalues α, α0. Let w be an eigenvector corresponding to α0. Then, with respect to v, w the matrix of B must have the form µ ¶ α b 12 . 0 β µ ¶ 0 0 with b 6= 0. It is now straightforward to verify that is the unique 12 0 1 invariant hermitean matrix. Moreover it is degenerate, which it should be as A, B have a common eigenvector. 2

F.Beukers: Hypergeometric Functions, preliminary notes 2.2. MONODROMY OF THE HYPERGEOMETRIC FUNCTION 29

Definition 2.2.9 With the assumptions as in the previous lemma let G be the group generated by A and B. Then G is called hyperbolic, euclidean, spheric if F is indefinite, degenerate, definite respectively. Corollary 2.2.10 Let {x} denote the fractional part of x (x minus largest integer ≤ x). Suppose that (4.1) is irreducible. Let F be the invariant hermitean form for the monodromy group. In particular, the sets {{a}, {b}} and {0, {c}} are disjoint. If {c} is between {a} and {b} then F is positive definite (spherical case). If {c} is not between {a} and {b} then F is indefinite (hyperbolic case). The most pittoresque way to describe the monodromy group is by using Schwarz’ triangles. First a little geometry. Definition 2.2.11 A curvilinear triangle is a connected open subset of C ∪ ∞ = P1 whose boundary is the union of three open segments of a circle or straight line and three points. The segments are called the edges of the triangles, the points are called the vertices. It is an exercise to prove that, given the vertices and the corresponding angles (< π), a curvilinear triangle exists and is uniquely determined This can be seen best by taking the vertices to be 0, 1, ∞. Then the edges connected to ∞ are actually straight lines. More generally, a curvilinear triangle in C ∪ ∞ = P1 is determined by its angles (in clockwise ordering) up to a Möbiustransformation. Let z0 be a point in the upper half plane H = {z ∈ C|=(z) > 0} and let f, g be two independent solutions of the hypergeometric equation near z0. The quotient D(z) = f/g, considered as a map from H to P1, is called the Schwarz map and we have the following picture and theorem.

D(1)

D(z)

D(∝)

0 1 ∝ D(0) Theorem 2.2.12 (Schwarz) Let λ = |1 − c|, µ = |c − a − b|, ν = |a − b| and Suppose 0 ≤ λ, µ, ν < 1. Then the map D(z) = f/g maps H ∪ R one-to-one onto a curvilinear triangle. The vertices correspond to the points D(0),D(1),D(∞) and the corresponding angles are λπ, µπ, νπ.

F.Beukers: Hypergeometric Functions, preliminary notes 30 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

As to the proof of Schwarz’ theorem, the following three ingredients are important. – The map D(z) is locally bijective in every point of H. Notice that D0(z) = (f 0g − fg0)/g2. The determinant f 0g − fg0 is the Wronskian determinant of our equation and equals z−c(1 − z)c−a−b−1. In particular it is non-zero in H. When g has a zero at some point z1 we simply consider 1/D(z) instead. Since f and g cannot vanish at the same time in a regular point, we have f(z1) 6= 0. – The map D(z) maps the segments (∞, 0), (0, 1), (1, ∞) to segments of circles or straight lines. For example, since a, b, c ∈ R we have two real solutions on (0, 1) (see Kummer’s solutions). Call them f,˜ g˜. Clearly, the function D˜(z) = f/˜ g˜ maps (0, 1) on a segment of R. Since f, g are C-linear combinations of f,˜ g˜ we see that D(z) is a M¨obiustransform of D˜(z). Hence D(z) maps (0, 1) to a segment of a circle or a straight line. – The map D(z) maps a small neighbourhood of 0 to a sector with angle |1−c| = λ and similarly for 1, ∞. This follows from the fact that near z = 0 the functions f, g are C-linear combinations of F (a, b, c|z) and z1−cF (a − c + 1, b − c + 1, 2 − c|z). For the exact determination of the image of the Schwarz map we need the following additional result.

Proposition 2.2.13 (Gauss) Suppose that a, b, c ∈ R, c 6∈ Z≤0 and c > a + b. Prove that Γ(c − a)Γ(c − b) F (a, b, c|1) = . Γ(c)Γ(c − a − b) This can be proven by evaluation of Euler’s integral using the Euler Beta-function. To study the analytic continuation of D(z) we use Schwarz’ reﬂection principle. Hopefully, the following picture illustrates how this works.

D(1)

D(z)

D(∝) 0 1 ∝ D(0)

F.Beukers: Hypergeometric Functions, preliminary notes 2.3. TRIANGLE GROUPS 31

The monodromy group modulo scalars arises as follows. Let W be the group generated by the reﬂections in the edges of the curvilinear triangle. The monodromy group is the subgroup of W consisting of all elements which are product of an even number of reﬂections. In the following section we shall study precisely such groups.

2.3 Triangle groups

In this section we let S be either the Poincar´edisk {z ∈ C| |z| < 1}, C or P1. equipped with the hyperbolic, euclidean and spherical metric respectively.

Deﬁnition 2.3.1 A (geodesic) triangle is an connected open subset of S, of ﬁnite volume, whose boundary in S is a union of three open segments of a geodesic and at most three points. The segments are called the edges of the triangles, the points are called the vertices.

We ﬁrst point out that under very mild conditions any curvilinear triangle can be thought of as a geodesic triangle.

Lemma 2.3.2 Let λ, µ, ν be real numbers in the interval [0, 1). There exists a geodesic triangle with angles λπ, µπ, νπ if and only if λ+µ+ν < 1+2 min(λ, µ, ν).

Proof. Suppose first that λ+µ+ν < 1. Our condition is then trivially satisfied. For any such curvilinear triangle we can take the common orthogonal circle of the three edges, which will become the boundary of a Poincarédisk. The edges are then automatically geodesics. Suppose that λ + µ + ν = 1. Our condition is equivalent to saying that all angles are positive. In this case geodesic triangles are planar triangles in the euclidean geometry with finite area. The latter property is equivalent to positivity of all angles. Suppose that λ + µ + ν > 1. From spherical geometry it follows that a spherical triangle exists if and only if our condition is satisfied. We let W (∆) be the group of isometries of S generated by the 3 reflections through the edges of a geodesic triangle ∆. First we look at subgroups generated by reflection in two intersecting geodesics.

Lemma 2.3.3 Let ρ, σ be two geodesics intersecting in a point P with an angle πλ. Let r, s be the reflections in ρ, σ respectively. Then the group D generated by r, s is a dihedral group consisting of rotations (rs)n around P with angles 2nπλ, n ∈ Z and reflections in the lines (rs)n(ρ), (rs)n(σ). In particular D is finite of order 2m if and only if λ = q/m for some q ∈ Z with q 6= 0 and gcd(m, q) = 1. Furthermore, D is discrete if and only if λ is either zero or a rational number.

F.Beukers: Hypergeometric Functions, preliminary notes 32 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

Theorem 2.3.4 For any geodesic triangle ∆ we have S = ∪γ∈W (∆)γ(∆), where ∆ denotes the closure of ∆ in S.

Proof. First of all we note that there exists a positive d0 with the following property. For any point P whose distance to ∆ is less than d0 there exists γ ∈ W (∆) such that P ∈ γ(∆). For γ we can simply take a suitable element from one of the dihedral reﬂection groups around the vertices. A fortiori, any point P with distance less than d0 from ∪γ∈W (∆)γ(∆) belongs to this set. As a consequence the set ∪γ∈W (∆)γ(∆) is open and closed in S, hence our theorem follows. 2

Deﬁnition 2.3.5 An elementary triangle is a geodesic triangle whose vertex angles are all of the form π/n, n ∈ Z≥2 ∪ ∞.

Theorem 2.3.6 Let ∆ be an elementary triangle. Then, for any γ ∈ W (∆), γ 6= Id we have γ(∆) ∩ ∆ = ∅.

Proof. This is a special case of the theorem of Coxeter-Tits on representations of Coxeter groups. See Humphreys book on Reﬂection groups and Coxeter groups [H]. A group G of isometries acting on S is said to act discretely if there exists a point P ∈ S and a positive d0 such that distance(P, g(P ))> d0 whenever g 6= Id. In particular it follows from the previous theorem that triangle groups generated by elementary triangles act discretely. The following theorem characterises all groups W (∆) which act discretely on the symmetric space S.

Theorem 2.3.7 Suppose W = W (∆) acts discretely. Then there exists an elementary triangle ∆el such that W (∆) = W (∆el). Moreover, ∆ is a ﬁnite union of copies of ∆el under elements of W .

Proof. First of all note that the vertex angles must be either 0 or rational mul- tiples of π, otherwise the corresponding dihedral group is not discrete. We shall show that if ∆ is not elementary, then there exists a geodesic triangle ∆0 such that W (∆) = W (∆0) and Vol(∆0) ≤ Vol(∆)/2. If ∆0 is not elementary we repeat the process and so on. However, there is a limit to these processes since, by discreteness, there is a positive lower bound to Vol(∆00) for any ∆00 satisfying 00 W (∆) = W (∆ ). Hence we must hit upon an elementary triangle ∆el such that W (∆) = W (∆el). Let α, β, γ be the edges of ∆ and rα, rβ, rγ the corresponding reﬂections. Suppose that the vertex angle between α and β is of the form mπ/n with gcd(m, n) = 1, but m > 1. Let δ be the geodesic between α and β whose angle with α is π/n. Let rδ be the reﬂection in δ. Then the dihedral group generated by rα and rβ is the

F.Beukers: Hypergeometric Functions, preliminary notes 2.3. TRIANGLE GROUPS 33

0 same as the one generated by rα and rδ. Let ∆ be the triangle with edges α, δ, γ. Then, clearly, W (∆) = W (∆0). If the volume of ∆0 is larger than half the volume of ∆ we simply perform the above construction with α and β interchanged. 2

Below we give a list of non-elementary triangles ∆ = (λ, µ, ν) with vertex angles λπ, µπ, νπ which allow a dissection with elementary triangles ∆el such that W (∆) = W (∆el). In the spherical case discreteness of W (∆) implies ﬁniteness. The list of spherical cases was already found by H.A.Schwarz and F.Klein (see [Kl]). In the following table N denotes the number of congruent elementary triangles needed to cover ∆.

λ µ ν N elementary 2/n 1/m 1/m 2 × (1/2, 1/n, 1/m) n odd 1/2 2/n 1/n 3 × (1/2, 1/3, 1/n) n odd 1/3 3/n 1/n 4 × (1/2, 1/3, 1/n) n 6≡ 0 mod 3 2/n 2/n 2/n 6 × (1/2, 1/3, 1/n) n odd 4/n 1/n 1/n 6 × (1/2, 1/3, 1/n) n odd 2/3 1/3 1/5 6 × (1/2, 1/3, 1/5) 1/2 2/3 1/5 7 × (1/2, 1/3, 1/5) 3/5 2/5 1/3 10 × (1/2, 1/3, 1/5) 1/3 2/7 1/7 10 × (1/2, 1/3, 1/7)

As an application we construct a hypergeometric function which is algebraic over C(z). Take the triangle (4/5, 1/5, 1/5), which is spherical. Corresponding values for a, b, c can be taken to be 1/10, −1/10, 1/5. Hence the quotient of any two solutions f, g of the corresponding hypergeometric is algebraic. Its derivative (f 0g − fg0)/g2 is algebraic and so is the Wronskian determinant f 0g − fg0 = z−c(1 − z)c−a−b−1. Hence g and, a fortiori, f are algebraic. In particular, F (1/10, −1/10, 1/5|z) is an algebraic function.

In many cases it is also possible to ﬁnd elementary triangles ∆el which can 0 be dissected into isometric copies of a smaller elementary triangle ∆el. Hence 0 W (∆el) ⊂ W (∆el). The most spectacular example is the dissection of the triangle (1/7, 1/7, 1/7) into 24 copies of (1/2, 1/3, 1/7). As a corollary of this dissection we ﬁnd the remarkable identity

µ ¯ ¶ µ ¯ ¶ 2 3 6¯ 1 29 6¯ z(z − 1)(z3 − 8z2 + 5 ∗ z + 1) F , , ¯ z = b(z)−1/28 F , , ¯ 123 2 1 7 7 7¯ 2 1 84 84 7¯ b(z)3 where b(z) = 1 − 236z + 1666z2 − 3360z3 + 3395z4 − 1736z5 + 42z6 + 228z7 + z8. For a complete list of such dissections and the corresponding identities we refer to [V].

F.Beukers: Hypergeometric Functions, preliminary notes 34 CHAPTER 2. GAUSS HYPERGEOMETRIC FUNCTIONS

2.4 Some loose ends

In the Schwarz map we have assumed that the parameters a, b, c are such that λ = |1 − c|, µ = |c − a − b|, ν = |a − b| are all less than 1. It turns out that in the irreducible case this is no restriction, since we can shift a, b, c by integers without aﬀecting the monodromy group. In fact,

Lemma 2.4.1 Assume that none of the numbers a, b, c−a, c−b is integral. There exist a0 ∈ a(mod Z), b0 ∈ b(mod Z), c0 ∈ c(mod Z) such that

0 ≤ λ, µ, ν < 1 λ + µ + ν < 1 + 2 min(λ, µ, ν) where λ = |1 − c0|, µ = |c0 − a0 − b0|, ν = |a0 − b0|. In the case λ + µ + ν < 1 there exists only one choice for a0, b0, c0 and in the case λ + µ + ν > 1 there exist four possible choices.

Proof. First of all let us suppose that 0 ≤ a, b, c < 1. Without loss of generality we can assume that a ≤ b. We consider the following cases. Case i) 0 < a < c < b < 1. We take a0 = a, b0 = b, c0 = c. Then, λ = 1 − c, µ = a + b − c, ν = b − a and the inequalities are satisfied. Moreover, λ + µ + ν = 1 + 2b − 2c > 1. Case ii) 0 < a ≤ b < c < 1. We take a0 = a, b0 = b, c0 = c. When c ≥ a + b we get λ = 1 − c, µ = c − a − b, ν = b − a and the inequalities hold. Moreover, λ + µ + ν = 1 − 2a < 1. When c ≤ a + b we get λ = 1 − c, µ = a + b − c, ν = b − a and the inequalities hold. Moreover, λ + µ + ν = 1 + 2b − 2c < 1. Case iii) 0 ≤ c < a ≤ b < 1. We take a0 = a, b0 = b, c0 = c + 1. Then, λ = c, µ = c + 1 − a − b, ν = b − a and the inequalities are readily verified. Moreover, λ + µ + ν = 1 + 2c − 2a < 1. As to uniqueness we note that an integral shift in the a, b, c such that the corresponding values of λ, µ, ν stay below 1 necessarily gives the substitutions of the form λ → 1 − λ, µ → 1 − µ, ν → ν and similar ones where two of the parameters are replaced by 1 minus their value. In casethe condition λ + µ + ν < 1 + 2 min(λ, µ, ν) is violated by such a substitution. For example, λ + µ + ν ≤ 1 implies 1 − λ + 1 − µ + ν = 2 − (λ + µ + ν) + 2ν ≥ 1 + 2ν. In the spherical case the condition is not violated. When we have obtained a geodesic Schwarz triangle in our construction we automatically have a metric which is invariant under the projective monodromy group. This closely reflects the nature of the natural hermitian form on the monodromy group itself.

Theorem 2.4.2 Let a, b, c ∈ R be such that

0 ≤ λ, µ, ν < 1 λ + µ + ν < 1 + 2 min(λ, µ, ν)

F.Beukers: Hypergeometric Functions, preliminary notes 2.4. SOME LOOSE ENDS 35 where λ = |1 − c|, µ = |c − a − b|, ν = |a − b|. Let M be the monodromy group of (4.1). Then, M is spheric ⇐⇒ λ + µ + ν > 1 M is euclidean ⇐⇒ λ + µ + ν = 1 M is hyperbolic ⇐⇒ λ + µ + ν < 1.

Proof. In the case when none of the numbers a, b, c − a, c − b is integral, this statement can already be inferred from the proof of the previous lemma (we get only the hyperbolic and spheric case). It remains to show that if one of the numbers a, b, c − a, c − b is integral, we have λ + µ + ν = 1. Let us suppose for example that a ∈ Z. Notice that |a − b| < 1 and |a + b| < |c| + 1 < 3. Hence |a| ≤ |a − b|/2 + |a + b|/2 < 2. So, a = 0, ±1. A case by case analysis using the inequalities for λ, µ, ν yields our statement. 2

F.Beukers: Hypergeometric Functions, preliminary notes Chapter 3

Generalised hypergeometric functions nFn−1

3.1 Deﬁnition, ﬁrst properties

Let α1, . . . , αn; β1, . . . , βn be any complex numbers and consider the generalised hypergeometric equation in one variable, d z(D + α ) ··· (D + α )F = (D + β − 1) ··· (D + β − 1)F,D = z (3.1) 1 n 1 n dz This is a Fuchsian equation of order n with singularities at 0, 1, ∞. The local exponents read,

1 − β1,..., 1 − βn at z = 0 α1, . . . , αn at z = ∞ Pn 0, 1, . . . , n − 2, −1 + 1 (βi − αi) at z = 1

When the βi are distinct modulo 1 a basis of solutions at z = 0 is given by the functions µ ¯ ¶ α1 − βi + 1, . . . , αn − βi + 1 ¯ z1−βi F ¯ z (i = 1, . . . , n). n n−1 ∨ ¯ β1 − βi + 1, .. .., βn − βi + 1

∨ Here .. .. denotes suppression of the term βi − βi + 1 and nFn−1 stands for the generalised hypergeometric function in one variable µ ¯ ¶ ¯ X∞ α1, . . . , αn ¯ (α1)k ··· (αn)k k nFn−1 ¯ z = z . β1, . . . , βn−1 (β1)k ··· (βn−1)kk! k=0 At z = 1 we have the following interesting situation. Theorem 3.1.1 (Pochhammer) The equation (4.1) has n−1 independent holomorphic solutions near z = 1.

36 3.2. MONODROMY 37

¡ ¢ d n The proof of this result follows from the observation that the coeﬃcient of dz in (4.1) equals zn+1 − zn and the following theorem.

Theorem 3.1.2 Consider the linear diﬀerential equation

(n) (n−1) 0 pn(z)y + pn−1(z)y + ··· + p1(z)y + p0(z)y = 0 where the pi(z) are analytic around a point z = a. Suppose that pn(z) has a zero of order one at z = a. The the diﬀerential equations has n − 1 independent holomorphic solutions around z = a.

Proof. Without loss of generality we can assume that a = 0. Write pi(z) = P j j≥0 pijz for every j. Then, in particular, pn0 = 0 and pn1 6= 0. we determine a P k power series solution k≥0 fkz by substituting it into the equation. We obtain the recursion relations,

Finally we mention the Euler integral for nFn−1(α1, . . . , αn; β1, . . . , βn−1|z),

n−1 Z Z Q Y 1 1 n−1 αi−1 βi−αi−1 Γ(βi) i=1 ti (1 − ti) ··· dt1 ··· dtn−1 Γ(α )Γ(β − α ) (1 − zt ··· t )αn i=1 i i i 0 0 1 n−1 for all <βi > <αi > 0 (i = 1, . . . , n − 1).

3.2 Monodromy

1 Fix a base point z0 ∈ P − {0, 1, ∞}, say z0 = 1/2. Denote by G the fundamental 1 group π1(P − {0, 1, ∞}). Clearly G is generated by the simple loops g0, g1, g∞ around the corresponding points together with the relation g0g1g∞ = 1. Let V (α, β) = V (α1, . . . , αn; β1, . . . , βn) be the local solution space of (4.1) around z0. Denote by M(α, β): G → GL(V (α, β)) the monodromy representation of (4.1). Write

h0 = M(α, β)g0 h1 = M(α, β)g1 h∞ = M(α, β)g∞.

The eigenvalues of h0 and h∞ read exp(−2πiβj) and exp(2πiαj) respectively. Since there are n−1 independent holomorphic solutions near z = 1 the element h1 has n−1 eigenvalues 1 together with n−1 independent eigenvectors. Equivalently, rank(h1 − Id) ≤ 1. An element h ∈ GL(V ) such that rank(h − Id) = 1 will be called a (pseudo)-reflection. The determinant of a reflection will be called the special eigenvalue. From the relation between the generators of the fundamental −1 group it follows that h1 = h∞h0 is a (pseudo)reflection.

F.Beukers: Hypergeometric Functions, preliminary notes 38 CHAPTER 3. HYPERGEOMETRIC FUNCTIONS N FN−1

Theorem 3.2.1 Let H ⊂ GL(n, C) be a subgroup generated by two matrices A, B such that AB−1 is a reﬂection. Then H acts irreducibly on Cn if and only if A and B have disjoint sets of eigenvalues.

Proof. Suppose that H acts reducibly. Let V1 be a nontrivial invariant subspace n and let V2 = C /V1. Since A − B has rank 1, A and B coincide on either V1 or V2. Hence they have a common eigenvalue. Suppose conversely that A and B have a common eigenvalue λ. Let W = ker(A− B). Since AB−1 − Id has rank one, the same holds for A − B. Hence dim(W ) = n − 1. If any eigenvector of A belongs to W , it must also be an eigenvector of B, since A and B coincide on W . Hence there is a one-dimensional invariant subspace. Suppose W does not contain any eigenvector of A or B. We show that the subspace U = (A − λ)Cn is invariant under H. Note that A − λId has a non-trivial kernel which has trivial intersection with W . Hence U has dimension n − 1 and U = (A − λ)W . Since A − λ and B − λ coincide on W we conclude that also U = (B − λ)W and hence, by a similar argument as for A, U = (B − λ)Cn. Notice that U is stable under A, as follows trivially from A(A − λ)Cn = (A − λ)ACn = (A − λ)Cn. For a similar reason U is stable under B. Hence H has the invariant subspace U. 2

Corollary 3.2.2 The monodromy group of (4.1) acts irreducibly if and only if all diﬀerences αi − βj are non-integral.

−1 This Corollary follows by application of our Theorem with A = h∞ and B = h0 . From now on we shall be interested in the irreducible case only.

∗ Theorem 3.2.3 (Levelt) Let a1, . . . , an; b1, . . . , bn ∈ C be such that ai 6= bj for all i, j. Then there exist A, B ∈ GL(n, C) with eigenvalues a1, . . . , an and −1 b1, . . . , bn respectively such that AB is a reﬂection. Moreover, the pair A, B is uniquely determined up to conjugation.

Proof. First we show the existence. Let Y n n−1 (X − ai) = X + A1X + ··· + An i Y n n−1 (X − bi) = X + B1X + ··· + Bn i and     0 0 ... 0 −An 0 0 ... 0 −Bn  1 0 ... 0 −A   1 0 ... 0 −B   n−1   n−1  A =  0 1 ... 0 −An−2  B =  0 1 ... 0 −Bn−2   . .   . .   . .   . .  0 0 ... 1 −A1 0 0 ... 1 −B1

F.Beukers: Hypergeometric Functions, preliminary notes 3.3. HYPERGEOMETRIC GROUPS 39

Then rank(A − B) = 1, hence rank(AB−1 − Id) = 1 and AB−1 is a reﬂection. To prove uniqueness of A, B we let W = ker(A − B). Note that dim W = n − 1. Let V = W ∩ A−1W ∩ · · · ∩ A−(n−2)W . Then dim V ≥ 1. Suppose dim V > 1. Choose v ∈ V ∩ A−(n−1)W . Then Aiv ∈ W for i = 0, 1, . . . , n − 1. Hence i U =< A v >i∈Z⊂ W is A-stable. In particular, W contains an eigenvector of A. Since B = A on W this is also an eigenvector of B with the same eigenvalue, contradicting our assumption on A, B. Hence dim V = 1. Letting v ∈ V we take v, Av, . . . , An−1v as basis of Cn. Since A = B on W we have that Aiv = Biv for i = 0, 1, . . . , n − 2 and with respect to this basis A and B have automatically the form given above. 2

Corollary 3.2.4 With the same hypotheses and Ai,Bj as in the proof of the previous theorem we have that < A, B > can be described by matrices having elements in Z[Ai,Bj, 1/An, 1/Bn].

Levelt’s theorem is a special case of a general rigidity theorem which has recently been proved by N.M.Katz. In the last section we shall give an elementary proof of Katz’s theorem.

3.3 Hypergeometric groups

∗ Deﬁnition 3.3.1 Let a1, . . . , an; b1, . . . , bn ∈ C . such that ai 6= bj for every i, j. The group generated by A, B such that A and B have eigenvalues a1, . . . , an −1 and b1, . . . , bn respectively and such that AB is a pseudoreﬂection, will be called a hypergeometric group with parameters ai and bj. Notation: H(a, b) = a1, . . . , an; b1, . . . , bn.

In particular, the monodromy group of (4.1) is a hypergeometric group with 2πiαk 2πiβk ak = e and bk = e .

Theorem 3.3.2 Let H be a hypergeometric group with parameters a1, . . . , an and b1, . . . , bn. Suppose that these parameters lie on the unit circle in C. Then there P n exists a non-degenerate hermitean form F (x, y) = Fijxiyj on C such that F (hx, hy) = F (x, y) for all h ∈ H and all x, y ∈ Cn. Denote by ≺, ¹ the total ordering on the unit circle corresponding to increasing argument. Assume that the a1 ¹ ... ¹ an and b1 ¹ ... ¹ bn. Let mj = #{k|bk ≺ aj} for j = 1, . . . , n. Then the signature (p, q) of the hermitean form F is given by ¯ ¯ ¯ n ¯ ¯X ¯ |p − q| = ¯ (−1)j+mj ¯ . ¯ ¯ j=1

F.Beukers: Hypergeometric Functions, preliminary notes 40 CHAPTER 3. HYPERGEOMETRIC FUNCTIONS N FN−1

Deﬁnition 3.3.3 Let a1, . . . , an and b1, . . . , bn be sets on the unit circle. We say that these sets interlace on the unit circle if and only if either

a1 ≺ b1 ≺ a2 ≺ b2 · · · ≺ an ≺ bn or

b1 ≺ a1 ≺ b2 ≺ a2 · · · ≺ bn ≺ an.

Corollary 3.3.4 Let the hypergeometric group H have all of its parameters on the unit circle. Then H is contained in U(n, C) if and only if the parametersets interlace on the unit circle.

Theorem 3.3.5 Suppose the parameters {a1, . . . , an} and {b1, . . . , bn} are roots of unity, let us say h-th roots of unity for some h ∈ Z≥2. Then the hypergeometric group H(a, b) is ﬁnite if and only if for each k ∈ Z with (h, k) = 1 the sets k k k k {a1, . . . , an} and {b1, . . . , bn} interlace on the unit circle.

Proof. The Galois group of Q(exp(2πi/h)) over Q is given by elements of the form

σk : exp(2πi/h) → exp(2πik/h) for any k, (k, h) = 1. The group H(a, b) can be represented by matrices with entries in the ring of cyclotomic integers Z[exp(2πi/h)]. The Galois automorphsim σk establishes an isomorphism between H(a, b) and the hypergeometric group Hk k k k k with parameters a1, . . . , an, b1, . . . , bn. Each group Hk has an invariant hermitian form Fk for (k, h) = 1. Suppose H(a, b) is finite. Then each Fk is definite, hence every pair of sets k k k k {a1, . . . , an} and {b1, . . . , bn} interlace on the unit circle. k k k k Suppose conversely that {a1, . . . , an} and {b1, . . . , bn} interlace for every k, (k, h) = 1. Then each group Hk is subgroup of a unitary group with definite form Fk. In particular the entries of each element are bounded in absolute value by some constant, C say. This implies that any entry of any element of H(a, b) has conju- gates which are all bounded by C. Since there exist only finitely many elements of Z[exp(2πi/h)] having this property, we conclude the finiteness of H(a, b). 2 An immediate consequence of this theorem is that, for example, the hypergeometric function µ ¯ ¶ 1/30, 7/30, 11/30, 13/30, 17/30, 19/30, 23/30, 29/30¯ F ¯ z 8 7 1/8, 1/4, 3/8, 1/2, 5/8, 3/4, 7/8 ¯ is an algebraic function.

F.Beukers: Hypergeometric Functions, preliminary notes 3.4. RIGIDITY 41

3.4 Rigidity

In this section we formulate and prove Katz’s result on rigidity, see [Ka]. Let k be a ﬁeld and g1, g2, . . . , gr ∈ GL(n, k) Let G be the group generated by g1, . . . , gr. We say that the r-tuple is irreducible if the group G acts irreducibly on kn. ˜ We call the r-tuple g1, . . . , gr linearly rigid if for any conjugatesg ˜1,..., hr of −1 g1, . . . , gr withg ˜1g˜2 ··· g˜r = Id there exists u ∈ GL(n, k) such thatg ˜i = ugiu for i = 1, 2, . . . , r.

For example, it follows from Levelt’s theorem that the generators g1 = A, g2 = −1 −1 B , g3 = BA of a hypergeometric group form a linearly rigid system.

Theorem 3.4.1 (Katz) Let g1, g2, . . . , gr ∈ GL(n, k) be an irreducible r-tuple with g1g2 . . . gr = Id. Let, for each i, δi be the codimension of the linear space {A ∈ Mn(k)|giA = Agi} (centralizer of gi). Then,

2 i) δ1 + ··· + δr ≥ 2(n − 1)

2 ii) If δ1 + ··· + δr = 2(n − 1), the system is linearly rigid.

iii) If k is algebraically closed, then the converse of part ii) holds

We note that the centraliser of g ∈ GL(n, k) depends only on the Jordan normal form of g. If g is diagonalisable, the dimension of the centraliser is the sum of the squares of the dimensions of the eigenspaces of g. When g has distinct eigenvalues this dimension is n, when g is a (pseudo)reﬂection this dimension is (n−1)2 +1 = n2 −2n+2. The corresponding codimensions are n2 −n and 2n−2 respectively.

By way of example consider a hypergeometric group generated by g1 = A, g2 = −1 −1 B , g3 = BA . In general A and B each have distinct eigenvalues, so δ1 = 2 δ2 = n − n. Since g3 is a (pseudo)reflection we have δ3 = 2n − 2. Notice that 2 −1 −1 δ1 +δ2 +δ3 = 2n −2. Hence the triple A, B ,BA is linearly rigid. As a bonus we get that the eigenspaces of A and B all have dimension one. Hence to each eigenvalue there is precisely one Jordan block in the Jordan normal form. Another example comes from the Jordan-Pochhammer equation, which is an n- th order Fuchsian equation with n + 1 singular points and around each singular point the local monodromy is (up to a scalar) a pseudo-reflection. So for each 2 singularity we have δi = 2n − 2. The sum of these delta’s is of course 2(n − 1). So if the monodromy is irreducible we have again a rigid system. This case has been elaborated by [Ha]. The proof of Katz’s theorem is based on the following Theorem from linear alge- bra. In this Theorem we consider a group G acting on a finite dimensional linear space V . For every X ⊂ G we denote by d(X) resp. d∗(X) the codimension of the common fixed point space in V resp. V ∗, the dual of V , of all elements of X.

F.Beukers: Hypergeometric Functions, preliminary notes 42 CHAPTER 3. HYPERGEOMETRIC FUNCTIONS N FN−1

Theorem 3.4.2 (L.L.Scott) Let H ∈ GL(V ) be the group generated by h1, h2, . . . , hr with h1h2 ··· hr = Id. Then

∗ d(h1) + d(h2) + ··· + d(hr) ≥ d(H) + d (H).

r Proof. Let W be the direct sum ⊕i=1(1−hi)V . Deﬁne the linear map β : V → W by β : v 7→ ((1 − h1)v, . . . , (1 − hr)v). Deﬁne the linear map δ : W → V by

δ :(v1, . . . , vr) 7→ v1 + h1v2 + h1h2v3 + ··· + h1 ··· hr−1vr

Because of the identity

1 − h1h2 ··· hr = (1 − h1) + h1(1 − h2) + ··· + h1 ··· hr−1(1 − hr) we see that the image of β is contained in the kernel of δ. Hence dim(=β) ≤ r dim(ker δ). Moreover, the kernel of β is precisely ∩i=1 ker(1 − hi). The dimension of the latter space equals n − d(H). Hence dim(=β) = n − (n − d(H)) = d(H). The image of δ is

(1 − h1)V + h1(1 − h2)V + ··· + h1 ··· hr−1(1 − hr)V which is equal to (1 − h1)V + (1 − h2)V + ··· + (1 − hr)V . Note that any w ∈ r ∗ ∗ ∗ ∩i=1 ker(1−hi ) in the dual space V we vanishes on =δ. Hence dim(=δ) ≥ d (H). Pr Finally notice that dim(W ) = i=1 d(hi). Putting everything together we get Xr d(hi) = dim(W ) = dim(ker δ) + dim(=δ) i=1 ≥ dim(=β) + dim(=δ) ≥ d(H) + d∗(H)

This is precisely the desired inequality. 2

Proof of Katz’s theorem. We follow the approach of V¨olklein-Strambach [VS]. For the ﬁrst part of Katz’s theorem we apply Scott’s Theorem to the vector space −1 of n × n-matrices and the group generated by the maps hi : A 7→ gi Agi. No- tice that d(hi) is now precisely the codimension of the centraliser of gi, hence d(hi) = δi for all i. The number d(H) is precisely the codimension of the space {A ∈ Mn(k)|gA = Ag for all g ∈ G}. By Schur’s Lemma the irreducibility of the action of G implies that the dimension of this space is 1 and the codimension n2 − 1. So d(H) = n2 − 1. To determine d∗(H) we note that the ∗ matrix space V = Mn(k) is isomorphic to its dual via the map V → V given by A 7→ (X 7→ Trace(AX). Let us identify V with V ∗ in this way. Since

F.Beukers: Hypergeometric Functions, preliminary notes 3.4. RIGIDITY 43

Trace(Ag−1Xg) = Trace(gAg−1X) we see that the action of g on the dual space is given by A 7→ gAg−1. hence d∗(H) = n2 − 1. Application of Scott’s Theorem now shows that

∗ 2 δ1 + ··· + δ2 ≥ d(H) + d (H) = 2(n − 1) To prove the second part of the theorem we apply Scott’s Theorem with V = −1 Mn(k) again, but now with the maps hi : A 7→ gi Ag˜i. For each i choose −1 ui ∈ GL(n, k) such thatg ˜i = uigiui . Now note that

−1 d(hi) = codim{A|gi Ag˜i = A}

= codim{A|Ag˜i = giA} −1 = codim{A|Auigiui = giA}

= codim{A|(Aui)gi = gi(Aui)}

= codim{A|Agi = giA} = δi

2 The sum of the δi is given to be 2(n − 1). Together with Scott’s Theorem this implies d(H) + d∗(H) ≤ 2(n2 − 1). This means that either d(H) < n2 or d∗(H) < n2 or both. Let us assume d(H) < n2, the other case being similar. Then there is a non-trivial n × n matrix A such that Ag˜i = giA for all i. From these inequalities we see in particular that the image of A is stable under the group generated by the gi. Since the r-tuple g1, . . . , gr is irreducible this means that A(kn) is either trivial or kn itself. Because A is non-trivial we conclude that n n −1 A(k ) = k and A is invertible. We thus conclude thatg ˜i = A giA for all i. In other words, our system g1, . . . , gr is rigid. The proof of part iii) uses a dimension argument. Let Ci be the conjugacy class of gi i = 1, 2, . . . , r. Consider the multiplication map Π : C1 × C2 × · · · × Cr → GL(n, k) given by (c1, c − 2, . . . , cr) 7→ c1c2 ··· cr. We have

−1 dim(C1 × · · · × Cr) ≤ dim(Π (Id) + dim(=Π)

Pr Pr First of all note that dim(C1 × · · · × Cr) = i=1 dim(Ci) = i=1 δi. Secondly, by the rigidity and irreducibility assumptions we have dim(Π−1(Id)) = n2 − 1. Finally, =Π is contained in the hypersuface of all matrices whose determinant is 2 det(g1g2 ··· gr) = 1. Hence dim(=Π) ≤ n − 1. Pr 2 These three facts imply that i=1 δi ≤ 2(n − 1). Together with part i) this implies the desired equality. 2 In many practical situations the local monodromies of diﬀerential equations have eigenvalues which are complex numbers with absolute value 1. In that case there exists also a monodromy invariant Hermitian form on the solution space. We formulate this as a Lemma.

Lemma 3.4.3 Let g1, g2, . . . , gr ∈ GL(n, C) be a rigid, irreducible system with t −1 g1g2 ··· gr = Id. Suppose that for each i the matrices gi and g˜i = (gi) are

F.Beukers: Hypergeometric Functions, preliminary notes 44 CHAPTER 3. HYPERGEOMETRIC FUNCTIONS N FN−1

t conjugate. Then there exists a non-trivial matrix H ∈ Mn(C) such that giHgi = t H for each i and H = H.

Proof. Notice that,g ˜1 ··· g˜r = Id. Moreover, the gi andg ˜i are conjugate so by −1 rigidity there exists a matrix H ∈ GL(n, C) such thatg ˜i = HgiH for all i. t Hence H = giHgi for all i. Moreover, since the system g1, . . . , gr is irreducible, t the matrix H is uniquely determined up to a scalar factor. Since H is also a t solution we see that H = λH for some λ ∈ C. Moreover |λ| = 1 and writing λ = µ/µ we see that µH is a Hermitian matrix. Now take H := µH. 2

F.Beukers: Hypergeometric Functions, preliminary notes Chapter 4

Explicit monodromy for hypergeometric equations

4.1 Introduction

Notice that the proof of Levelt’s Theorem 3.2.3 provides us a very explicit construction of the monodromy matrices of the hypergeometric equation. However, the basis with respect to which these monodromy matrices occur do appear in the proof. So Levelt’s Theorem gives us only a determination of the monodromy group up to conjugation. In many applications it is desirable to have the explicit matrices with respect to an explicitly given basis of solutions. This is precisely the purpose of this chapter. We consider the hypergeometric equation d z(D + α ) ··· (D + α )F = (D + β − 1) ··· (D + β − 1)F,D = z . (4.1) 1 n 1 n dz We consider it in the complex plane with the positive real axis deleted. That is, all complex z with | arg(−z)| < π. We fix a basis of local solutions at z = 0 and at z = ∞. We continue the local solutions around 0 analytically in C − R≥0 to ∞ and compare the continued solutions with the local solutions at ∞. The coefficients that occur enable us to compute the desired monodromy matrices. To simplify matters we assume that the local bases of solutions do not contain logarithms, that is the parameters αi are distinct modulo 1 and the parameters βi are distinct modulo 1. A solution basis around z = 0 can then be given by µ ¯ ¶ α1 − βi + 1, . . . , αn − βi + 1 ¯ (−z)1−βi F ¯ z (i = 1, . . . , n) n n−1 ∨ ¯ β1 − βi + 1, .. .., βn − βi + 1 ∨ where the sign denotes suppression of βi − βi + 1. To make the final formulas more elegant we multiply this solution with the constant

Γ(α1 − βi + 1) ··· Γ(αn − βi + 1)

Γ(β1 − βi + 1) ··· Γ(βn − βi + 1)

45 46 CHAPTER 4. EXPLICIT MONODROMY and get the solution X Γ(α − β + k + 1) ··· Γ(α − β + k + 1) 1−βi 1 i n i k Fi := (²z) z Γ(β1 − βi + k + 1) ··· Γ(βn − βi + k + 1) k≥0 where we introduced the extra factor ² = (−1)n−1 for reasons that will become clear later. We also agree that for the determination of (²z)1−βi we choose −(n + 2)π < arg(²z) < −nπ. Similarly, around z = ∞ a basis can be given by X Γ(α − β + k + 1) ··· Γ(α − β + k + 1) −αi i 1 i n k Gi = (²z) (1/z) Γ(αi − α1 + k + 1) ··· Γ(αi − αn + k + 1) k≥0 In order to determine the connection between these solution sets we propose to use the technique of Mellin-Barnes integrals.

4.2 Mellin-Barnes integrals

In [GM] Golyshev anf Mellit describe a way to determine explicit monodromy of hypergeometric functions by studying Fourier transforms of products of factors of the form 1/Γ(γ ± s). In this section we adopt an approach inspired by them and which gives precisely the same formulas, namely Mellin-Branes type integrals of products of factors Γ(γ ± s). Let αi, βj ∈ C for i, j = 1, 2, . . . , n and suppose from now on that the αi, βj are all distinct modulo 1. This means we can write explicit solution bases for the hypergeometric equations as in the previous section and, by Corollary 3.2.2, the monodromy representation is irreducible. Let Γ be a path in the complex plane from i∞ to −i∞ and which bends in such a way that all points −αi − k (i = 1, . . . , n, k ∈ Z≥0) are on the left of Γ and all points −βi +k+1 (i = 1, . . . , n, k ∈ Z≥0) are on the right of Γ. Let i ∈ {1, . . . , n} and consider the integral Z 1 s I = Γ(α1 + s) ··· Γ(αn + s)Γ(1 − β1 − s) ··· Γ(1 − βn − s)(²z) ds 2πi Γ where ² = (−1)n−1. From Stirling’s formula (see [AAR, p21]) it follows that when s = a + bi and a1 < a < a2 and |b| → ∞ that √ |Γ(a + bi)| = 2π|b|a−1/2e−π|b|/2[1 + O(1/|b|)].

Notice also that |(−z)a+bi| = |(²z)a|e−b arg(²z)| for all a, b ∈ R. Putting these estimates together we see that the integral converges absolutely for all z ∈ C with | arg(²z)| < nπ. The interval (−nπ, nπ) can be subdivided into n intervals of the form ((2r−n−2)π, (2r−n)π). So, depending on the choice of determination s of (²z) the integral I represents n functions Ir on C \ R≥0 indexed by r. We now

F.Beukers: Hypergeometric Functions, preliminary notes 4.2. MELLIN-BARNES INTEGRALS 47

compute Ir as a power series in z. So we assume |z| < 1 and (2r − n − 2)π < arg(²z) < (2r−n)π. For any real a we denote by Γa the vertical path from a+i∞ to a−i∞. Let a0 be larger than all real parts of the −αi. We deform the contour

Γ to Γa0 such that all points −αi − k with k ∈ Z≥0 stay on the left. Then we continue to shift Γa0 to the right via the paths Γa with a → ∞. In the process the deformed paths may pass through a pole of Γ(1 − β1 − s) ··· Γ(1 − βn − s) and no others. That is, the points 1 − βi, 2 − βi,... for i = 1, . . . , n. The residue of the integrand of Ir at the pole s = k + 1 − βi equals

k 1−βi (²z) (²z) Γ(α1−βi+k+1) ··· Γ(αn−βi+k+1)Γ(−β1+βi−k) ··· Γ(−βn+βi−k) where the factor Γ(−βi + βi − k) is to be read as 1/k!. Once again we apply the identity Γ(x)Γ(1 − x) = sin πx to obtain Yn Y Γ(αl − βi + k + 1) π (²z)k(²z)1−βi Γ(βl − βi + k + 1) sin π(−βl + βi − k) l=1 l6=i where the factor sin π(−βi + βi − k) is omitted. So we get

Xn Y π X Yn Γ(α − β + k + 1) 1−βi l i k Ir = Ii,a + (²z) z sin π(−βl + βi) Γ(βl − βi + k + 1) i=1 l6=i k l=1 where the summation is over k = 0, 1, 2,... ba+<(βi)c and Ii,a denotes integration over Γa. Finally we note that |Ii,a| → 0 as a → ∞, simply because |z| < 1 and so |(−z)a+bi| decreases exponentially in a as a → ∞. Therefore we conclude that Xn Fi I = πn−1 e−2πirβi Q r = 1, 2, . . . , n. r sin π(−β + β ) i=1 l6=i l i

We now compute Ir as a power zeries in 1/z. So we assume |z| > 1 and (2r − n − 2)π < arg(²z) < (2r − n)π. Let a0 be a real number smaller than all real parts of the −βi. We now deform the contour Γ to Γa0 while keeping all points

−βi + k + 1 with k ∈ Z≥0 on the right. From then on we shift Γa0 to Γa where we let a → −∞. In the process the deformed paths may pass through the poles of Γ(s + α1) ··· Γ(s + αn) and no others. That is, the points −αj − k with k ∈ Z≥0. Use the fact that the residue of Γ(x) at x = k with k ≤ 0 is given by (−1)k/k!. We obtain that the residue of the integrand equals

−k −αj (²z) (²z) Γ(α1−αj−k) ··· Γ(αn−αj−k)Γ(−β1+αj+k+1) ··· Γ(−βn+αj+k+1) where the factor Γ(αj − αj − k) is to be read as 1/k!. We use the identity Γ(x)Γ(1 − x) = π/ sin πx once again. We get Yn Y Γ(αj − β1 + k + 1) π (²)−k(²z)−αj . Γ(αj − α1 + k + 1) sin(π(αl − αj − k)) l=1 l6=i

F.Beukers: Hypergeometric Functions, preliminary notes 48 CHAPTER 4. EXPLICIT MONODROMY

a+bi The integral over Γa tends to zero as a → −∞ because |z| > 1 and |(−z) | tends exponentially to 0 as a → −∞. Hence we conclude that for |z| > 1 we have Xn Gj I = πn−1 e−2πirαj Q , r = 1, 2, . . . , n. r sin(π(α − α )) j=1 l6=j l j There is an interesting consequence.

Corollary 4.2.1 Let notation be as above. With respect to the basis of solutions I1,...,In the monodromy matrix around z = 0 reads   0 1 0 ··· 0    0 0 1 ··· 0   . . .   . . .   0 0 0 ··· 1  −Bn −Bn−1 Bn−2 · · · −B1

n n−1 −2πiβk where X +B1X +···+Bn−1X +Bn is the polynomial with zeros e , k = 1, . . . , n. Similarly, around the point z = ∞ the monodromy matrix with respect to I reads   0 1 0 ··· 0    0 0 1 ··· 0   . . .   . . .   0 0 0 ··· 1  −An −An−1 An−2 · · · −A1

n n−1 −2πiαk where X +A1X +···+An−1X +An is the polynomial with zeros e , k = 1, . . . , n.

So we see that we have found an explicit basis of solutions of the hypergeometric equation with respect to which the monodromy has the shape given in Levelt’s Theorem 3.2.3.

t Proof. Let us denote I = (I1,...,In) and denote by F the vector with coordi- nates πn−1F Q i . l6=i sin π(−βl + βi)

We have seen above that I = MβF where Mβ is the VanderMonde type matrix   e−2πiβ1 e−2πiβ2 ··· e−2πiβn  e−4πiβ1 e−4πiβ2 ··· e−4πiβn   . . .  .  . . .  e−2nπiβ1 e−2nπiβ2 ··· e−2nπiβn A closed loop around z = 0 in positive direction gievs the local monodromy −2πiβj F → DβF where Dβ is the diagonal matrix with entries e . Hence the

F.Beukers: Hypergeometric Functions, preliminary notes 4.2. MELLIN-BARNES INTEGRALS 49

−1 solutions I goes over into MβIβF = MβIβMβ I. The local monodromy matrix with respect to I reads   0 1 0 ··· 0    0 0 1 ··· 0  −1  . . .  MβIβMβ =  . . .  .  0 0 0 ··· 1  −Bn −Bn−1 Bn−2 · · · −B1

The calculation around z = ∞ runs similarly.

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F.Beukers: Hypergeometric Functions, preliminary notes 50 CHAPTER 4. EXPLICIT MONODROMY

[V ] R.Vidunas, Transformations of Gauss hypergeometric functions, J.Computational and Applied Math 178 (2005), 473-487, arxiv: math.CA/0310436.

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F.Beukers: Hypergeometric Functions, preliminary notes